{"id":2491,"date":"2016-09-18T22:38:54","date_gmt":"2016-09-18T22:38:54","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=2491"},"modified":"2024-08-21T14:23:17","modified_gmt":"2024-08-21T14:23:17","slug":"resolucao-comentada-dos-exercicios-de-fisica-sobre-estrutura-atomica-atomo-de-bohr","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/fisica-moderna\/estrutura-atomica-atomo-de-bohr\/resolucao-comentada-dos-exercicios-de-fisica-sobre-estrutura-atomica-atomo-de-bohr\/","title":{"rendered":"\u00c1tomo de Bohr – Estrutura At\u00f4mica"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de f\u00edsica sobre<\/b><\/span><\/span><\/span> <\/b><\/p>\n

Estrutura At\u00f4mica \u2013 \u00c1tomo de Bohr<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

01-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Quando o el\u00e9tron salta novamente para sua \u00f3rbita original, de n\u00edveis de energia mais baixos, o que faz com que a energia seja liberada atrav\u00e9s da emiss\u00e3o de um f\u00f3ton luminoso.\u00a0 —\u00a0\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span> <\/b>R- D\u00a0\u00a0—\u00a0 veja teoria<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>I. Falsa\u00a0 —\u00a0 n\u00e3o \u00e9 cont\u00ednua, s\u00f3 pode ter valores determinados.<\/b><\/span><\/span><\/span><\/p>\n

II. Correta.<\/b><\/span><\/span><\/span><\/p>\n

III. Correta.<\/b><\/span><\/span><\/span><\/p>\n

R- D<\/b><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u0394E=h.f=h.c\/\u03bb\u00a0 —\u00a0 3,4.10<\/b><\/span><\/span><\/span>-19<\/b><\/span><\/span><\/sup><\/span>=6,6.10<\/b><\/span><\/span><\/span>-34<\/b><\/span><\/span><\/sup><\/span>.3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>\/\u03bb\u00a0 —\u00a0 \u03bb=5.824.10<\/b><\/span><\/span><\/span>-10<\/b><\/span><\/span><\/sup><\/span>\u00a0m\u00a0 —\u00a0\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe que para saltar do n\u00edvel 1 (n<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>) para o n\u00edvel zero (n<\/b><\/span><\/span><\/span>\u221e<\/b><\/span><\/span><\/sub><\/span>), onde o el\u00e9tron se ioniza, foram gastos (E<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0– E<\/b><\/span><\/span><\/span>\u221e<\/b><\/span><\/span><\/sub><\/span>)=\u2502-13,6 – 0\u2502= 13,6 eV\u00a0 —\u00a0 como recebeu 20 eV e gastou 16,3 eV, v\u00e3o sobrar\u00a0 —\u00a0 20,0 \u2013 13,6=6,4 eV\u00a0 —\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span> <\/b>R- C\u00a0 —\u00a0 veja teoria<\/b><\/span><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Lembre-se de que a velocidade de todas as cores e de todas as radia\u00e7\u00f5es eletromagn\u00e9ticas \u00e9 sempre a mesma no v\u00e1cuo e, aproximadamente no ar (3,0.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>m\/s)\u00a0 —\u00a0\u00a0 cada cor tem velocidades diferentes em outros meios homog\u00eaneos e transparentes (vidro, \u00e1gua, etc.) —\u00a0 a energia de cada cor \u00e9 diretamente proporcional \u00e0 freq\u00fc\u00eancia (f) e inversamente proporcional ao comprimento de onda (\u03bb)\u00a0 —\u00a0 da tabela, o menor \u03bb, a mais energ\u00e9tica \u00e9 a violeta\u00a0 —\u00a0\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Na transi\u00e7\u00e3o de 2 para 1\u00a0 —\u00a0 \u0394E=(-3,4) \u2013 (-13,6)\u00a0 —\u00a0\u00a0\u00a0 \u0394E=10,2 eV\u00a0 —\u00a0 compat\u00edvel com f<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>, que ser\u00e1 absorvido\u00a0 —\u00a0 De 3 para 1\u00a0 —\u00a0\u00a0\u00a0 \u0394E=(-1,5) \u2013 (-13,6)\u00a0 —\u00a0\u00a0\u00a0 \u0394E=12,1 eV \u00a0—\u00a0 compat\u00edvel com f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>, que ser\u00e1 absorvido\u00a0 —\u00a0 de 3 para 2\u00a0 —\u00a0\u00a0\u00a0 \u0394E=(-1,5) \u2013 (3,4)\u00a0 —\u00a0 \u0394E=1,9 eV\u00a0 —\u00a0 incompat\u00edvel com f<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>, que n\u00e3o ser\u00e1 absorvido\u00a0 —\u00a0\u00a0R- f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0e f<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Quando a transi\u00e7\u00e3o ocorrer de um n\u00edvel mais alto de energia para um n\u00edvel mais baixo (\u00f3rbita mais externa para \u00f3rbita mais interna), ocorre a emiss\u00e3o de um f\u00f3ton\u00a0 —\u00a0 como ao menor comprimento de onda corresponde a maior freq\u00fc\u00eancia e consequentemente maior n\u00edvel energ\u00e9tico voc\u00ea deve escolher a alternativa em que ocorre maior varia\u00e7\u00e3o de energia\u00a0 —\u00a0\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>\u2502\u0394E\u2502<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>= hc\/\u03bb=4,13.10<\/b><\/span><\/span><\/span>-15<\/b><\/span><\/span><\/sup><\/span>.3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>\/1,03.10<\/b><\/span><\/span><\/span>-7<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 \u2502\u0394E\u2502<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>=12.03 eV\u00a0 —\u00a0 \u2502\u0394E\u2502<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>= hc\/\u03bb= 4,13.10<\/b><\/span><\/span><\/span>-15<\/b><\/span><\/span><\/sup><\/span>.3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>\/4,85.10<\/b><\/span><\/span><\/span>-7<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 \u2502\u0394E\u2502<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=<\/b><\/span><\/span><\/span><\/p>\n

2,55 eV\u00a0 —\u00a0 os valores mais pr\u00f3ximos s\u00e3o de 2 e 6\u00a0 —\u00a0 \u2502\u0394E\u2502<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>= 13,6 \u2013 1,51=12,09\u00a0 —\u00a0 \u2502\u0394E\u2502<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sub><\/span>=3,40 \u2013 0,85=2,55\u00a0 —\u00a0 R- B<\/b><\/span><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u0394E=E<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=(-3,4) \u2013 (-13,6)\u00a0 —\u00a0 \u0394E=10,2 eV=10,2.1,6.10<\/b><\/span><\/span><\/span>-19<\/b><\/span><\/span><\/sup><\/span>\u00a0 — \u0394E= 16,3.10<\/b><\/span><\/span><\/span>-19<\/b><\/span><\/span><\/sup><\/span>J\u00a0 —\u00a0 \u0394E=h.f\u00a0 —\u00a0 f=16,3.10<\/b><\/span><\/span><\/span>-19<\/b><\/span><\/span><\/sup><\/span>\/6,63.10<\/b><\/span><\/span><\/span>-34<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0f=2,5.10<\/b><\/span><\/span><\/span>15<\/b><\/span><\/span><\/sup><\/span>Hz<\/b><\/span><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Pelo gr\u00e1fico, a intensidade \u00e9 m\u00e1xima quando f=1,8.10<\/b><\/span><\/span><\/span>13<\/b><\/span><\/span><\/sup><\/span>\u00a0Hz\u00a0 —\u00a0 c=\u03bbf\u00a0 —\u00a0 3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>=\u03bb.1,8.10<\/b><\/span><\/span><\/span>13<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0\u03bb=1,7.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>m<\/b><\/span><\/span><\/span><\/p>\n

b) Estimando a \u00e1rea, considere a pessoa como um cilindro de 0,3m de di\u00e2metro e altura 1,7m\u00a0 —\u00a0 A=2\u03c0Rh + 2\u03c0R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=2.3.0,15.1,7 + 2.3.(0,15)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 A\u22481,7m<\/b><\/span><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup><\/span>\u00a0—\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=E\/\u0394t\u00a0 —\u00a0 \u03c3<\/b><\/span><\/span><\/span>\uf020<\/b><\/span><\/span><\/span>A(T<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 T<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>a<\/b><\/span><\/span><\/sub><\/span>) = E\/9.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 6.10<\/b><\/span><\/span><\/span>-8<\/b><\/span><\/span><\/sup><\/span>.1,7.((37 + 273)<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>\u00a0 – (27 + 273)<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>)=E\/9.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

10,2.10<\/b><\/span><\/span><\/span>-8<\/b><\/span><\/span><\/sup><\/span>.(310<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>\u00a0– 300<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>)=E\/9.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 E=10,2.10<\/b><\/span><\/span><\/span>-8<\/b><\/span><\/span><\/sup><\/span>.9.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>.(9.235.210.000 \u2013 8.100.000.000)\u00a0 —\u00a0\u00a0E\u22481,04.10<\/b><\/span><\/span><\/span>7\u00a0<\/b><\/span><\/span><\/sup><\/span>J<\/b><\/span><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Se os f\u00f3tons foram emitidos, o el\u00e9tron pulou da \u00f3rbita 2 para a 1\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>= -3,40 \u2013 ( -13,6)\u00a0 —\u00a0 \u0394E= + 10,2 eV\u00a0 —\u00a0\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O exerc\u00edcio est\u00e1 afirmando que a freq\u00fc\u00eancia do vermelho \u00e9 menor que a do verde\u00a0 —\u00a0 a freq\u00fc\u00eancia \u00e9 diretamente proporcional ao n\u00edvel energ\u00e9tico\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>vermelho\u00a0<\/b><\/span><\/span><\/sub><\/span>< E<\/b><\/span><\/span><\/span>verde<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 a freq\u00fc\u00eancia \u00e9 inversamente proporcional ao comprimento de onda \u03bb e consequentemente ao n\u00edvel energ\u00e9tico E\u00a0 —\u00a0 \u03bb<\/b><\/span><\/span><\/span>vermelho<\/b><\/span><\/span><\/sub><\/span>\u00a0> \u03bb<\/b><\/span><\/span><\/span>verde<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Para o \u00e1tomo de hidrog\u00eanio\u00a0 —\u00a0 (E<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>)=- 13,6(1\/(n<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0– (n<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>)\u00a0 —\u00a0 \u0394E= – 13,6.(1\/16 \u2013 1\/1)\u00a0 —\u00a0 \u0394E= – 13,6.(-15\/16)\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\u0394E= +12,75 eV\u00a0\u00a0—\u00a0 ou\u00a0 —\u00a0\u00a0\u0394E=12,75.1,6.10<\/b><\/span><\/span><\/span>-19<\/b><\/span><\/span><\/sup><\/span>=20,4.10<\/b><\/span><\/span><\/span>-19<\/b><\/span><\/span><\/sup><\/span>J\u00a0 —\u00a0 \u0394E=h.f\u00a0 —\u00a0 20,4.10<\/b><\/span><\/span><\/span>-19<\/b><\/span><\/span><\/sup><\/span>=6,63.10<\/b><\/span><\/span><\/span>-34<\/b><\/span><\/span><\/sup><\/span>.f\u00a0 —\u00a0\u00a0f=3,07.10<\/b><\/span><\/span><\/span>15<\/b><\/span><\/span><\/sup><\/span>\u00a0Hz<\/b><\/span><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Nas radiografias os \u00a0ossos saem brancos e tecidos em volta negros\u00a0 —\u00a0 isso ocorre porque o osso, cuja estrutura \u00e9 mais densa que a do tecido mole, absorve mais radia\u00e7\u00e3o ficando com apar\u00eancia clara enquanto que o tecido mole, menos denso, \u00e9 atravessado pelos raios X, ficando com apar\u00eancia mais escura\u00a0 —\u00a0\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Lembre-se de que a energia dos f\u00f3tons de luz \u00e9 diretamente proporcional \u00e0 frequ\u00eancia da luz f\u00a0 —\u00a0\u00a0 como a freq\u00fc\u00eancia da luz amarela, do s\u00f3dio, \u00e9 menor, ent\u00e3o sua energia tamb\u00e9m \u00e9 menor\u00a0 —\u00a0 a velocidade de propaga\u00e7\u00e3o de todas as cores e de todas as radia\u00e7\u00f5es eletromagn\u00e9ticas \u00e9 sempre a mesma no ar e no v\u00e1cuo (c=3,0.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>m\/s)\u00a0 —\u00a0\u00a0R- A<\/b><\/span><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Basta substituir na equa\u00e7\u00e3o, que \u00e9 fornecida, lembrando que a velocidade v \u00e9 c (velocidade da luz)\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=2GM\/R\u00a0 —\u00a0 R=2GM\/c<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00a0—\u00a0 R=2.6,67.10<\/b><\/span><\/span><\/span>-11<\/b><\/span><\/span><\/sup><\/span>.5,98.10<\/b><\/span><\/span><\/span>24<\/b><\/span><\/span><\/sup><\/span>(3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 R\u22489.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>m\u00a0 —\u00a0 di\u00e2metro\u00a0d=2R=2.9.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>=18.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>=0,018m=1,8cm<\/b><\/span><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Fiss\u00e3o Nuclear\u00a0 —\u00a0 rea\u00e7\u00e3o que se inicia com o choque de um n\u00eautron com um n\u00facleo inst\u00e1vel que proporciona a quebra deste \u00faltimo e, por este motivo, \u00e9 chamado de fiss\u00e3o nuclear\u00a0(divis\u00e3o do n\u00facleo)\u00a0 —\u00a0 Fus\u00e3o Nuclear – \u00e9 o processo no qual dois ou mais n\u00facleos at\u00f4micos se juntam e formam um outro n\u00facleo de maior n\u00famero at\u00f4mico.<\/b><\/span><\/span><\/span>\"\"<\/p>\n

\"\"<\/p>\n

O principal tipo de fus\u00e3o que ocorre no interior das estrelas \u00e9 o de Hidrog\u00eanio em H\u00e9lio, onde dois pr\u00f3tons se fundem em uma part\u00edcula alfa (um n\u00facleo de h\u00e9lio), liberando dois p\u00f3sitrons, dois neutrinos e energia.<\/b><\/span><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span><\/span><\/p>\n

R- B<\/b><\/span><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span><\/span> <\/b>R- D\u00a0\u00a0—\u00a0 veja teoria<\/b><\/span><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 n=1\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=-E<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/1<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=-E<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 n=4\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span>=-E<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/4<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/16\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span>=(-E<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>) \u2013 (-E<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/16)\u00a0 —\u00a0 \u0394E<\/b><\/span><\/span><\/span>1,4<\/b><\/span><\/span><\/sub><\/span>=(-15\/16).E<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe no gr\u00e1fico 1 a queda de intensidade de radia\u00e7\u00e3o quando os raios X atravessam a parte maci\u00e7a do cilindro em teste\u00a0 —\u00a0 no gr\u00e1fico 2 voc\u00ea observa que a metade esquerda do cilindro, em compara\u00e7\u00e3o com o gr\u00e1fico 1, deve ser maci\u00e7a e a metade direita oca\u00a0 —\u00a0 R- E<\/b><\/span><\/span><\/span><\/p>\n

24-<\/b><\/span><\/span><\/span> <\/b>R- A\u00a0\u00a0—\u00a0 veja teoria<\/b><\/span><\/span><\/span><\/p>\n

25-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A for\u00e7a nuclear forte \u00e9 a respons\u00e1vel pela atra\u00e7\u00e3o m\u00fatua entre os n\u00eautrons e pr\u00f3tons do n\u00facleo at\u00f4mico\u00a0 —\u00a0 a for\u00e7a nuclear fraca participa das transmuta\u00e7\u00f5es at\u00f4micas\u00a0 —\u00a0 a for\u00e7a el\u00e9trica est\u00e1 associada aos campos el\u00e9tricos criados em torno das cargas el\u00e9tricas<\/b><\/span><\/span><\/span><\/p>\n

\u00a0 —\u00a0 a for\u00e7a gravitacional \u00e9 uma das mais presentes em nosso cotidiano, haja vista que nosso peso \u00e9 uma manifesta\u00e7\u00e3o gravitacional\u00a0 —\u00a0\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

26-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Velocidade da luz no v\u00e1cuo\u00a0 —\u00a0 \u00a0c = 3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>\u00a0m\/s\u00a0 —\u00a0 \u00a0constante de Planck\u00a0 —\u00a0 \u00a0h = 6,6.10<\/b><\/span><\/span><\/span>-34<\/b><\/span><\/span><\/sup><\/span>\u00a0J\u00b7s = 4,1.10<\/b><\/span><\/span><\/span>-15<\/b><\/span><\/span><\/sup><\/span>\u00a0eV\u00b7s\u00a0\u00a0 —\u00a0\u00a0 combinando a equa\u00e7\u00e3o fundamental da ondulat\u00f3ria com a equa\u00e7\u00e3o de Planck\u00a0 —\u00a0 E=h.c\/\u03bb\u00a0 —\u00a0 para a linha H\u03b1\u00a0 —\u00a0 \u03bb=656,3.10<\/b><\/span><\/span><\/span>-9<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00a0—\u00a0 E=4,1.10<\/b><\/span><\/span><\/span>-15<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>\/653,6.10<\/b><\/span><\/span><\/span>-9<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 E\u22481,88 eV\u00a0 —\u00a0\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

27-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>De acordo com o postulado de Bohr, o el\u00e9tron excitado passa para um n\u00edvel mais energ\u00e9tico\u00a0 —\u00a0 Ao sofrer decaimento para o n\u00edvel estacion\u00e1rio, ele emite um f\u00f3ton, que dependendo da frequ\u00eancia, poder\u00e1 ser ou n\u00e3o na forma de luz vis\u00edvel\u00a0 —\u00a0\u00a0R- A\u00a0 \u00a0<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

28-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>(I) Incorreta\u00a0 —\u00a0 \u00a0intensidade de radia\u00e7\u00e3o emitida aumenta com a temperatura\u00a0 —\u00a0\u00a0 I=\u03c3.T<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 portanto, T<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0> T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

(II) Correta\u00a0 —\u00a0 pelo gr\u00e1fico<\/b><\/span><\/span><\/span><\/p>\n

(III) Incorreta\u00a0 —\u00a0 pelo gr\u00e1fico, vemos que o comprimento de onda para o qual a radia\u00e7\u00e3o \u00e9 m\u00e1xima \u00e9 menor para T<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

(IV) Incorreta. J\u00e1 justificado em (I).<\/b><\/span><\/span><\/span><\/p>\n

(V) Incorreta. J\u00e1 justificado em (I). \u00a0<\/b><\/span><\/span><\/span><\/p>\n

R- E\u00a0<\/b><\/span><\/span><\/span><\/p>\n

29-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A lei de Stefan-Boltzmann afirma que a pot\u00eancia total irradiada pelo corpo negro \u00e9 diretamente proporcional \u00e0 \u00e1rea (S) da superf\u00edcie emissora e diretamente proporcional \u00e0 quarta pot\u00eancia da temperatura absoluta (T): P = sST<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 a equa\u00e7\u00e3o do efeito fotoel\u00e9trico dada por Einstein afirma que quando uma onda eletromagn\u00e9tica de alta frequ\u00eancia atinge uma chapa met\u00e1lica, cada f\u00f3ton pode arrancar um \u00fanico el\u00e9tron que \u00e9 ejetado com energia cin\u00e9tica m\u00e1xima (K) dada pela express\u00e3o\u00a0 —\u00a0 K = h f \u2013 W, sendo h a constante de Planck, f a frequ\u00eancia da onda incidente e W o trabalho para arrancar o el\u00e9tron do metal\u00a0 —\u00a0 a equa\u00e7\u00e3o de Louis de Broglie concilia as caracter\u00edsticas ondulat\u00f3rias e corpusculares dos fen\u00f4menos relacionados \u00e0 luz, atrav\u00e9s da equa\u00e7\u00e3o\u00a0 —\u00a0 \u03bb=h\/Q, sendo l o comprimento de onda associado ao movimento da part\u00edcula que se desloca com quantidade de movimento (momento linear) Q\u00a0 —\u00a0\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

30-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Os f\u00f3tons incidentes que podem ser absorvidos s\u00e3o determinados a partir das diferen\u00e7as de energia entre os estados inicial (n\u00edvel fundamental) e final (1\u00ba, 2\u00ba ou 3\u00ba n\u00edvel)\u00a0 —\u00a0 aqueles f\u00f3tons cujas energias coincidem com uma das diferen\u00e7as de energia entre os n\u00edveis mostrados na figura poder\u00e3o ser absorvidos\u00a0 —\u00a0 as diferen\u00e7as de energia entre os estados inicial e final s\u00e3o dadas por\u00a0 —\u00a0 DEf<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0= – 0,85 – ( -13,60) = 12,75eV (n\u00edvel fundamental e 3<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0n\u00edvel)\u00a0 —\u00a0 DEf<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0= – 1,51 – ( -13,60) = 12,09eV (n\u00edvel fundamental e 2<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>n\u00edvel)\u00a0 —\u00a0 DEf<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0= – 3,40 – ( -13,60) = 10,20eV (n\u00edvel fundamental e 1<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0n\u00edvel)\u00a0 —\u00a0 logo, os f\u00f3tons que podem ser absorvidos pelo \u00e1tomo de hidrog\u00eanio no estado fundamental s\u00e3o aqueles cujas energias s\u00e3o respectivamente iguais a: 12,09 eV, quando o \u00e1tomo \u00e9 excitado do estado fundamental para o 2\u00ba n\u00edvel, e 10,20 eV, quando o \u00e1tomo \u00e9 excitado do estado fundamental para o 1\u00ba n\u00edvel.<\/b><\/span><\/span><\/span><\/p>\n

b)<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

c) As energias dos f\u00f3tons emitidos s\u00e3o determinadas a partir da diferen\u00e7a de energia entre o n\u00edvel inicial e o n\u00edvel final\u00a0 —\u00a0 portanto, os f\u00f3tons emitidos ter\u00e3o as seguintes energias\u00a0 —\u00a0 DE2f = -1,51 – ( -13,60) = 12,09eV\u00a0 —\u00a0 DE1f = -3,40 – ( -13,60) = 10,20eV\u00a0 —\u00a0 DE21 = -1,51 – ( – 3,40) = 1,89eV<\/b><\/span><\/span><\/span><\/p>\n

31-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>1. Tomando os n\u00edveis 0, 1 e 2 de energias E<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>, E<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0e E<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>, tal que E<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0< E<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0< E<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0(ver representa\u00e7\u00e3o na figura acima) para transi\u00e7\u00f5es do n\u00edvel 2 para o 0, temos duas hip\u00f3tese\u00a0 —\u00a0 1\u00aa hip\u00f3tese\u00a0 —\u00a0 \u00a0emiss\u00e3o de 1 f\u00f3ton de energia E = h . f \u2013 E<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>o\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00a0—\u00a0 2\u00aa hip\u00f3tese\u00a0 —\u00a0 \u00a0emiss\u00e3o de 2 f\u00f3tons de energias E\u2019 = h . f\u2019 = E<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0e E\u201d = h . f\u201d = E<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 por conserva\u00e7\u00e3o da energia\u00a0 —\u00a0 \u00a0(E<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>) = (E<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>) + E<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 E = E\u2019 + E\u2019 e assim, h . f = h . f\u2019 + h . f\u201d\u00a0 —\u00a0 f = f\u2019 + f\u201d\u00a0 —\u00a0 portanto, cada frequ\u00eancia pode ser a soma ou a diferen\u00e7a entre outras.<\/b><\/span><\/span><\/span><\/p>\n

2. Os comprimentos de onda que limitam a regi\u00e3o do vis\u00edvel s\u00e3o l = 4,0.10<\/b><\/span><\/span><\/span>-7<\/b><\/span><\/span><\/sup><\/span>\u00a0m e\u00a0l\u2019 = 7,0.10<\/b><\/span><\/span><\/span>-7<\/b><\/span><\/span><\/sup><\/span>m (tabela de constantes), que correspondem, respectivamente, \u00e0s frequ\u00eancias f=3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>\/4.10<\/b><\/span><\/span><\/span>-7<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0f=7,5.10<\/b><\/span><\/span><\/span>14<\/b><\/span><\/span><\/sup><\/span>\u00a0Hz\u00a0 —\u00a0 f\u2019=3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>\/7.10<\/b><\/span><\/span><\/span>-7<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0f\u2019=4,3.10<\/b><\/span><\/span><\/span>14<\/b><\/span><\/span><\/sup><\/span>\u00a0Hz<\/b><\/span><\/span><\/span><\/p>\n

32-<\/b><\/span><\/span><\/span> <\/b>R- A\u00a0\u00a0—\u00a0 veja teoria<\/b><\/span><\/span><\/span><\/p>\n

33-<\/b><\/span><\/span><\/span> <\/b>R- B\u00a0 —\u00a0 veja teoria<\/b><\/span><\/span><\/span><\/p>\n

34-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) E=h.c\/ \u03bb\u00a0 —\u00a0 1,98.10<\/b><\/span><\/span><\/span>-15<\/b><\/span><\/span><\/sup><\/span>=6,6.10<\/b><\/span><\/span><\/span>-34<\/b><\/span><\/span><\/sup><\/span>.3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>\/\u03bb\u00a0 —\u00a0 \u03bb=19,8.10<\/b><\/span><\/span><\/span>-26<\/b><\/span><\/span><\/sup><\/span>\/1,98.10<\/b><\/span><\/span><\/span>-15<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0\u03bb=10<\/b><\/span><\/span><\/span>-10<\/b><\/span><\/span><\/sup><\/span>m\u00a0 —\u00a0 \u03bb=1 A<\/b><\/span><\/span><\/span><\/p>\n

b) Q \u2013 quantidade de movimento\u00a0 —\u00a0\u00a0 \u03bb=h\/mv\u00a0 —\u00a0 \u03bb=h\/Q\u00a0 —\u00a0 Q=h\/\u03bb=6,6.10<\/b><\/span><\/span><\/span>-34<\/b><\/span><\/span><\/sup><\/span>\/10<\/b><\/span><\/span><\/span>-10<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0Q=6,6.10<\/b><\/span><\/span><\/span>-24<\/b><\/span><\/span><\/sup><\/span>\u00a0kg.m\/s<\/b><\/span><\/span><\/span><\/p>\n

35-<\/b><\/span><\/span><\/span> <\/b>R- D<\/b><\/span><\/span><\/span><\/p>\n

36-<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R- A<\/b><\/span><\/span><\/span><\/p>\n

37-<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R- A<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

38-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe que na rea\u00e7\u00e3o para a forma\u00e7\u00e3o de uma mol\u00e9cula de glicose surgem seis mol\u00e9culas de CO<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 energia necess\u00e1ria\u00a0 —\u00a0 E=6.2,34.10<\/b><\/span><\/span><\/span>-18<\/b><\/span><\/span><\/sup><\/span>J\u00a0 —\u00a0 E=1,404.10<\/b><\/span><\/span><\/span>-17<\/b><\/span><\/span><\/sup><\/span>J\u00a0 —\u00a0 freq\u00fc\u00eancia f da radia\u00e7\u00e3o eletromagn\u00e9tica\u00a0 —\u00a0 c=\u03bbf\u00a0 —\u00a0 3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>=6,80.10<\/b><\/span><\/span><\/span>-7<\/b><\/span><\/span><\/sup><\/span>f\u00a0 —\u00a0 f=4,4.10<\/b><\/span><\/span><\/span>14<\/b><\/span><\/span><\/sup><\/span>Hz\u00a0 —\u00a0 equa\u00e7\u00e3o de Planck\u00a0 —\u00a0 E=n.h.f\u00a0 —\u00a0 n \u2013 n\u00famero de f\u00f3tons absorvidos\u00a0 —\u00a0 1,404.10<\/b><\/span><\/span><\/span>-7<\/b><\/span><\/span><\/sup><\/span>=n.6,62.10<\/b><\/span><\/span><\/span>-34<\/b><\/span><\/span><\/sup><\/span>.4,4.10<\/b><\/span><\/span><\/span>14<\/b><\/span><\/span><\/sup><\/span>\u00a0 — \u00a0n\u224848\u00a0 —\u00a0\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

39-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe que o pedido \u00e9 a raz\u00e3o R entre a intensidade do campo magn\u00e9tico no CMS (B<\/b><\/span><\/span><\/span>CMS<\/b><\/span><\/span><\/sub><\/span>) e do campo magn\u00e9tico da Terra (B<\/b><\/span><\/span><\/span>Terra<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 n=B<\/b><\/span><\/span><\/span>CMS<\/b><\/span><\/span><\/sub><\/span>\/B<\/b><\/span><\/span><\/span>Terra<\/b><\/span><\/span><\/sub><\/span>=4\/30.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 n\u2248133.333 vezes\u00a0 —\u00a0\u00a0R- A<\/b><\/span><\/span><\/span><\/p>\n

40-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Dado\u00a0 — \u00a0E<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0= 13,6 eV\u00a0 —\u00a0 pela conserva\u00e7\u00e3o da energia, a energia (E) do f\u00f3ton emitido \u00e9 em m\u00f3dulo, igual \u00e0 varia\u00e7\u00e3o da energia do el\u00e9tron\u00a0 —\u00a0 E=\u2502E<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/2<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/1<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u2502=\u250213,6\/4 \u2013 13,6\u2502\u00a0 —\u00a0\u00a0E=10,2 e<\/b><\/span><\/span><\/span><\/p>\n

b) Dados\u00a0 —\u00a0\u00a0 E = 10,2 eV; 1 eV = 1,6.10<\/b><\/span><\/span><\/span>-19<\/b><\/span><\/span><\/sup><\/span>\u00a0J\u00a0 —\u00a0 M<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>\u00a0= 1,6.10<\/b><\/span><\/span><\/span>-27<\/b><\/span><\/span><\/sup><\/span>\u00a0kg\u00a0 —\u00a0\u00a0 c = 3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>\u00a0m\/s\u00a0 —\u00a0 Q=E\/c\u00a0 —\u00a0 convers\u00e3o de el\u00e9tron-volt para joule\u00a0 —\u00a0 E = 10,2 (1,6.10<\/b><\/span><\/span><\/span>-19<\/b><\/span><\/span><\/sup><\/span>) J\u00a0 —\u00a0\u00a0pela conserva\u00e7\u00e3o da quantidade de movimento, o pr\u00f3ton adquire quantidade de movimento de mesma intensidade que o f\u00f3ton, em sentido oposto\u00a0 —\u00a0 assim, sendo v a velocidade adquirida pelo pr\u00f3ton\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

\u2502Q<\/b><\/span><\/span><\/span>f\u00f3ton<\/b><\/span><\/span><\/sub><\/span>\u2502= \u2502Q<\/b><\/span><\/span><\/span>pr\u00f3ton<\/b><\/span><\/span><\/sub><\/span>\u2502 —\u00a0 E\/c=M<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>.V\u00a0 —\u00a0 V=E\/(M<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>.c)\u00a0 —\u00a0 V=10,2.1,6.10<\/b><\/span><\/span><\/span>-19<\/b><\/span><\/span><\/sup><\/span>\/1,6.10<\/b><\/span><\/span><\/span>-27<\/b><\/span><\/span><\/sup><\/span>.3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0V=3,4m\/s<\/b><\/span><\/span><\/span><\/p>\n

41-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) n<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>=c\/v<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 1,3=3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>\/V<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0V<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>\u2248 2,3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>m\/s<\/b><\/span><\/span><\/span><\/p>\n

b)<\/b><\/span><\/span><\/span>\"\"<\/p>\n

Observe na figura\u00a0 —\u00a0 cos50<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=d\/d<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 0,64=1,6\/d<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=2,5m\u00a0 —\u00a0 sendo a velocidade da radia\u00e7\u00e3o Cerenkov constante\u00a0 — V<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=\u0394S\/\u0394t\u00a0 —\u00a0\u00a0 V<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=d<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>\/\u0394t\u00a0 —\u00a0 a radia\u00e7\u00e3o Cerenkov percorre a dist\u00e2ncia d<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>\u00a0no mesmo intervalo de tempo em que a luz percorre a dist\u00e2ncia d nesse meio\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=2,5\/12.10<\/b><\/span><\/span><\/span>-9<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0V<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>\u22482,1.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>m\/s<\/b><\/span><\/span><\/span><\/p>\n

42-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe a figura, que mostra o el\u00e9tron movendo-se em uma de suas \u00f3rbitas, sendo\u00a0 —\u00a0 m \u2013 massa do el\u00e9tron\u00a0 —\u00a0 e \u2013 carga elementar \u2013 m\u00f3dulo da carga do pr\u00f3ton que \u00e9 igual ao m\u00f3dulo da carga do el\u00e9tron=e\u00a0 —\u00a0 e \u2013 raio da \u00f3rbita do el\u00e9tron\u00a0 —\u00a0 k \u2013 constante eletrost\u00e1tica\u00a0 —\u00a0 a for\u00e7a eletrost\u00e1tica (\"\")\u00a0 trocada entre o pr\u00f3ton e o el\u00e9tron age como resultante centr\u00edpeta (\"\")\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/r=k.e.e\/r<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=ke<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/r\u00a0 —\u00a0 multiplicando ambos os membros por \u00bd\u00a0 —\u00a0 mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 = ke<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2r (I)\u00a0 —\u00a0 a energia (E) do el\u00e9tron \u00e9 a soma da sua energia cin\u00e9tica (E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>) com a sua energia potencial (E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 E=E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>\u00a0(II)\u00a0 —\u00a0 com o referencial no infinito\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>= – ke<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/r (III)\u00a0 —\u00a0 substituindo (I) e (III) em (II)\u00a0 —\u00a0 E=ke<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2r \u2013 ke<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/r\u00a0 —\u00a0 E= – ke<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2r (IV)\u00a0 —\u00a0 de acordo com o enunciado,<\/b><\/span><\/span><\/span>\"\"<\/p>\n

\u00a0o comprimento (C) da \u00f3rbita \u00e9 igual a um n\u00famero inteiro (n) de comprimentos de onda (\u03bb) do el\u00e9tron\u00a0 —\u00a0 C=n\u03bb\u00a0 —\u00a0 2\u03c0r=n\u03bb\u00a0 —\u00a0 r=n\u03bb\/2\u03c0 (V)\u00a0 —\u00a0 substituindo (V) em (IV)\u00a0 —\u00a0 E= – ke<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2(n\u03bb\/2\u03c0)\u00a0 —\u00a0\u00a0E= – k\u03c0e<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/n\u03bb<\/b><\/span><\/span><\/span><\/p>\n

43-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Dados\u00a0 —\u00a0 carga do pr\u00f3ton\u00a0 —\u00a0 q<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=e\u00a0 —\u00a0 carga do quark u\u00a0 —\u00a0 q<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span>=2\/3e \u00a0—\u00a0 calculando a carga do quark down\u00a0 —\u00a0 do enunciado\u00a0 —\u00a0 p=u + u + d\u00a0 —\u00a0\u00a0 q<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=2q<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span>\u00a0+ q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0 e=2.(2\/3)e + q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 e=(4\/3)e + q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>= – (1\/3)e\u00a0 —\u00a0 considere que o n\u00eautron seja formado de x quarks up e y quarks down\u00a0 — \u00a0como sua carga \u00e9 nula\u00a0 —\u00a0 x + y = 3\u00a0 —\u00a0 x(2\/3)e + y(-1\/3)e = 0\u00a0 —\u00a0 2x \u2013 y = 0\u00a0 —\u00a0 y=2x\u00a0 — X + 2x=3\u00a0 —\u00a0 x=1\u00a0 —\u00a0 y=2\u00a0 —\u00a0 conclui-se que\u00a0um n\u00eautron \u00e9 formado de 1 quark up e 2 quarks down (n=udd)<\/b><\/span><\/span><\/span>\"\"<\/p>\n

\u00a0<\/span><\/p>\n

44-<\/b><\/span><\/span><\/span> <\/b>a) Falsa\u00a0 — \u00a0A fiss\u00e3o nuclear \u2014 divis\u00e3o de um n\u00facleo at\u00f4mico pesado em dois n\u00facleos mais leves.<\/b><\/span><\/span><\/span><\/p>\n

b) Correta\u00a0 —\u00a0\u00a0 Durante a fus\u00e3o nuclear ocorre grande libera\u00e7\u00e3o de energia, j\u00e1 que as massas dos n\u00facleos produzidos s\u00e3o inferiores as dos n\u00facleos iniciais. Parte da massa perdida durante a fus\u00e3o nuclear \u00e9 convertida em energia, de acordo com a Equa\u00e7\u00e3o de Einstein E=MC<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>. Estas fus\u00f5es nucleares explicam o calor e a luz do Sol, percebidos por n\u00f3s, aqui na Terra.<\/b><\/span><\/span><\/span><\/p>\n

c) Falsa\u00a0 —\u00a0 veja b.<\/b><\/span><\/span><\/p>\n

d) Falsa\u00a0 —\u00a0 veja b.<\/b><\/span><\/span><\/span><\/p>\n

e) Falsa\u00a0 —\u00a0 veja b.<\/b><\/span><\/span><\/span><\/p>\n

R- B.\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

45-<\/b><\/span><\/span><\/span><\/p>\n

\u00a0Veja no esquema abaixo a soma das massas dos reagentes e a soma das massas dos produtos da rea\u00e7\u00e3o\u00a0 —\u00a0 diferen\u00e7a entre a soma dessas massas\u00a0 —\u00a0 \u2206m=236,05 \u2013 235,88=0,17 uma\u00a0 —\u00a0 equa\u00e7\u00e3o de Einstein de equival\u00eancia entre massa e energia\u00a0 —\u00a0 E= \u2206m.c<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 E=0,17uma=0,17.930MeV\u00a0 —\u00a0\u00a0E=158,1 MeV. (essa energia \u00e9 liberada principalmente sob forma de energia cin\u00e9tica).<\/b><\/span><\/span><\/span>\"\"<\/p>\n

\u2022 Decaimento alfa (\u03b1) \u2013 emiss\u00e3o de part\u00edculas \u03b1 (n\u00facleo de\u00a0<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>He)\u00a0 —\u00a0 decaimento beta (\u03b2) \u2013 emiss\u00e3o, pelo n\u00facleo, de part\u00edculas \u03b2 (cargas negativas, el\u00e9trons)\u00a0 —\u00a0 decaimento gama (\u03b3) \u2013 emiss\u00e3o de raios gama (f\u00f3tons de alta energia).<\/b><\/span><\/span><\/span><\/p>\n

46-<\/b><\/span><\/span><\/span> <\/b>a) Do texto\u00a0 —\u00a0 \u201cSegundo todos os c\u00e1lculos, as futuras usinas de fus\u00e3o nuclear poder\u00e3o extrair de 1 metro c\u00fabico de \u00e1gua uma quantidade de energia igual \u00e0 de 2 mil barris de petr\u00f3leo\u201d\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 1 m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 2.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0barris\u00a0 —\u00a0 100 m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 n barris\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

n=2.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 n=2.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0barris de petr\u00f3leo\u00a0 —\u00a0 como cada barril cont\u00e9m 1,5.10<\/b><\/span><\/span><\/span>6\u00a0<\/b><\/span><\/span><\/sup><\/span>kcal, 2.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0barris conter\u00e3o\u00a0 —\u00a0 W=2.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0barris x<\/b><\/span><\/span><\/span><\/p>\n

1,5.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>kcal\/barril\u00a0 —\u00a0 W=3,0.10<\/b><\/span><\/span><\/span>11<\/b><\/span><\/span><\/sup><\/span>\u00a0kcal=3,0.10<\/b><\/span><\/span><\/span>14<\/b><\/span><\/span><\/sup><\/span>cal\u00a0 —\u00a0 do enunciado\u00a0 —\u00a0 1 BEP (Barril Equivalente de Petr\u00f3leo), equivale a 1,45.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>\u00a0cal\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 1 BEP \u2013 1,45.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>\u00a0cal\u00a0 —\u00a0 n\u2019 BEP \u2013 3.10<\/b><\/span><\/span><\/span>14<\/b><\/span><\/span><\/sup><\/span>\u00a0cal\u00a0 —\u00a0 n\u2019=3.10<\/b><\/span><\/span><\/span>14<\/b><\/span><\/span><\/sup><\/span>\u00a0cal\/1,45.10<\/b><\/span><\/span><\/span>9\u00a0<\/b><\/span><\/span><\/sup><\/span>cal\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

n\u2019\u22482,07.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0BEP.<\/b><\/span><\/span><\/span><\/p>\n

b) Do texto: \u201cOs centros dos n\u00facleos dos \u00e1tomos de hidrog\u00eanio devem estar a 1.10<\/b><\/span><\/span><\/span>-15<\/b><\/span><\/span><\/sup><\/span>\u00a0metros um do outro para que ocorra a fus\u00e3o\u201d\u00a0 —\u00a0 ainda do texto\u00a0 —\u00a0 \u201cessa fus\u00e3o \u00e9 o processo no qual dois n\u00facleos de \u00e1tomos leves (por exemplo, o hidrog\u00eanio \u2013 cujo n\u00facleo \u00e9 constitu\u00eddo por 1 pr\u00f3ton com carga el\u00e9trica elementar \u00e9 1,6.10<\/b><\/span><\/span><\/span>-19<\/b><\/span><\/span><\/sup><\/span>C) se combinam, ou se fundem, constituindo um elemento mais pesado. Os n\u00facleos, ent\u00e3o, carregados positivamente, devem se aproximar suficientemente um do outro, ou seja, vencer a for\u00e7a de repuls\u00e3o eletrost\u00e1tica entre eles\u201d\u00a0 —\u00a0 portanto, s\u00e3o dados\u00a0 —\u00a0 d=1.10<\/b><\/span><\/span><\/span>-15<\/b><\/span><\/span><\/sup><\/span>m\u00a0 —\u00a0 .|Q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>|= |Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>|=1,6.10<\/b><\/span><\/span><\/span>-19<\/b><\/span><\/span><\/sup><\/span>C\u00a0 —\u00a0 k=9.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>N.m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/C<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 Lei de Coulomb\u00a0 —\u00a0\u00a0 F=k.|Q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>|.\u00a0<\/b><\/span><\/span><\/span>|Q<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub><\/span>|.\/d<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0= 9.10<\/b><\/span><\/span><\/span><\/span>9<\/b><\/span><\/span><\/span><\/sup><\/span>.1,6.10<\/b><\/span><\/span><\/span><\/span>-19<\/b><\/span><\/span><\/span><\/sup><\/span>.1,6.10<\/b><\/span><\/span><\/span><\/span>-19<\/b><\/span><\/span><\/span><\/sup><\/span>\/(1.10<\/b><\/span><\/span><\/span><\/span>-15<\/b><\/span><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0\u00a0 —\u00a0 F=23,04.10<\/b><\/span><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0F=230,4 N.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\u00a0<\/span>47-\u00a0<\/b><\/span><\/span><\/span><\/span><\/p>\n

As duas leis estabelecidas por Phillip Lenard para o efeito fotoel\u00e9trico s\u00e3o:<\/b><\/span><\/span><\/span>\"\"<\/p>\n

1\u00aa Lei: para determinada frequ\u00eancia, o n\u00famero de el\u00e9trons emitidos (conhecidos como fotoel\u00e9trons) pela placa iluminada (emissora)\u00a0<\/b><\/span><\/span><\/span>\u00e9 proporcional \u00e0 intensidade da luz<\/b><\/span><\/span><\/span><\/span>\u00a0nela incidente.<\/b><\/span><\/span><\/span><\/p>\n

2\u00aa Lei: a energia cin\u00e9tica dos fotoel\u00e9trons\u00a0<\/b><\/span><\/span><\/span>depende da frequ\u00eancia<\/b><\/span><\/span><\/span><\/span>\u00a0da radia\u00e7\u00e3o incidente na placa emissora, n\u00e3o dependendo da intensidade dessa radia\u00e7\u00e3o.<\/b><\/span><\/span><\/span><\/p>\n

Como, na situa\u00e7\u00e3o\u00a0b, a intensidade da luz incidente na placa emissora \u00e9 o dobro em rela\u00e7\u00e3o \u00e0 situa\u00e7\u00e3o\u00a0a, de acordo com a 1\u00aa lei, o n\u00famero de el\u00e9trons liberados tamb\u00e9m \u00e9 o dobro, provocando corrente el\u00e9trica tamb\u00e9m duas vezes maior\u00a0 —\u00a0 i<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>\u00a0= 2<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>i<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 como se trata da mesma radia\u00e7\u00e3o nas duas situa\u00e7\u00f5es (mesma frequ\u00eancia), a 2\u00aa lei garante que energia cin\u00e9tica dos fotoel\u00e9trons tamb\u00e9m \u00e9 a mesma, exigindo o mesmo potencial el\u00e9trico de frenamento (-V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

\u00a0Voltar para os exerc\u00edcios<\/span><\/a>
\n<\/span><\/h3>\n

\u00a0<\/span><\/p>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de f\u00edsica sobre Estrutura At\u00f4mica \u2013 \u00c1tomo de Bohr \u00a0 01-\u00a0Quando o el\u00e9tron salta novamente para sua \u00f3rbita original, de n\u00edveis de energia mais baixos, o que faz com que a energia seja liberada atrav\u00e9s da emiss\u00e3o de um f\u00f3ton luminoso.\u00a0 —\u00a0\u00a0R- B 02- R- D\u00a0\u00a0—\u00a0 veja teoria 03-\u00a0I. Falsa\u00a0 —\u00a0 n\u00e3o \u00e9 cont\u00ednua, s\u00f3 pode<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":2487,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"class_list":["post-2491","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2491","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/comments?post=2491"}],"version-history":[{"count":4,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2491\/revisions"}],"predecessor-version":[{"id":10725,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2491\/revisions\/10725"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2487"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=2491"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}