An electron in hydrogen atom first jumps form second excited state of first excited state and then from first excited state of ground state. Let the ratio of wavelength momentum and energy of photons emitted in these two cases be a,b and c respctively . Then

Options

A. `z = 1//x`
B. `x = 9//4`
C. `y = 5//27`
D. `z = 5//27`

Answer

B

Explanation

`E_(1) = (hc)/(lambda) [(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]`
For second excited state to first excited state
`E_(1) = (hc)/(lambda) [1/4 - 1/9] rArr (hc)/(lambda) ((5)/(36))`
For first excited state to ground excited state
`E_(2) = (hc)/(lambda) [1-(1)/(4)] rArr (hc)/(lambda) [(3)/(4)]`
(A) `(E_(1))/(E_(2)) = (5)/(27)` , (B) `(E_(1))/(E_(2)) = ((hc)/(lambda_(1)))((lambda_(2))/(hc)) = (lambda_(2))/(lambda_(1)) = 27/5`
(C) `P prop 1/(lambda) rArr (P_(1))/(P_(2)) = 5/27`