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When a hydrogen atom is excited from ground state to first excited state, thenA. its kinetic energy increases by 10.2 eVB. its kinetic energy decreases by 10.2 eVC. its potential energy increases by 20.4 eVD. its angular momentum increases by `1.05 xx 10^(-34) J-s` |
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Answer» Correct Answer - B::C::D ground state `n = 1` first excited state `n = 2` `KE = (1)/(4pi epsilon_(0)) (z = 1)` `KE = (14.4 xx 10^(-10))/(2r) eV` `(KE) = (14.4 xx 10^(-10))/(2 xx 0.53 xx 10^(-10)) eV = 13.58 eV` `(KE)_(2) = eV = 3.39 eV` KE decreases by `= 10.2 eV` Now `PE = (-1)/(4pi epsilon_(0)) (e^(2))/(r ) = (-14.4 xx 10^(-10))/(r ) eV` `(PE)_(1) = (-14.4 xx 10^(-10))/(0.53 xx 10^(-10)) eV = -27.1 eV` `(PE)_(2) = (-14.4 xx 10^(-10))/(0.53 xx 10^(-10) xx 4) = -6.79eV` Pe increased by `= 20 4 eV` Now Angular momentum , `L = mvr = (nh)/(2pi)` `L_(2) - L_(1) = (h)/(2pi) = (6.6 xx 10^(-34))/(6.28)` `1.05 xx 10^(-34) J - sec` |
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