When a hydrogen atom is excited from ground state to first excited state, then

Options

A. its kinetic energy increased by `20 eV`
B. its kinetic energy dereased by `10.2 eV`
C. its potential energy increased by `20.4 eV`
D. its angular momentum increased by `1.05 xx 10^(-34) J s`

Answer

B, C, D

Explanation

Ground state `n = 1`
First excited state `n = 2`
`KE = (1)/(4 pi epsilon_(0)) (e^(2))/(2 r) (z = 1)`
`KE = (14.4 xx 10^(-10))/(2 r) eV`
Now `r = 0.53 n^(2) Å (z = 1)`
`(KE)_(1) = (14.4 xx 10^(-10))/(2 xx 0.53 xx 10^(-10)) eV = 13.58 eV`
`(KE)_(2) = (14.4 xx 10^(-10))/(2 xx 0.53 xx 10^(-10) xx 4) eV = 3.39 eV`
`KE` decrease by `10.2 eV`
Now `PE = (-1)/(4 pi 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) eV = - 6.79 eV`
`PE` increase by `20.4 eV`
Now, angular momentum, `L = mvr= (nh)/(2 pi)`
`L_(2) - L_(1) = (nh)/(2 pi) = (6.6 xx 10^(-34))/(6.28) = 1.05 xx 10^(-34) J s`