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The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [`a_0` is Bohr radius] :A. ` h^2/( 4 pi ^2 ma_0^2)`B. ` h^2/( 16pi ^2 ma_0^2)`C. ` h^2/( 32 pi ^2 ma_0^2)`D. ` h^2/(6 4 pi ^2 ma_0^2)` |
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Answer» Correct Answer - C For Bohr orbit . ` K.E, = 1/2 mu^2` and `mur_n = (nh)/(2pi) ro u = (nh)/( 2pi mr_n)` ` :. K.. 1/2 m = (n^2 h^2 )/( 4pi^2 m^2 r_n^2) = (h^2 h^2)/(8 pi^2 mr_n^2)` For (II) Bohr orbit , n=2 `:. r_n = a_0 xx n^2 =4a_0 (a_0 = "Bohr radius" )` Now ` K.E. = ((2)^2 h^2)/(( 8 ppi^2 m 9 4 a_0)^2) = h^2/(32 pi^2 ma_0^2)`. |
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