An electron of stationary hydrogen atom jumps from 4th energy level to ground level. The velocity that the photon acquired as a result of electron transition will be (where, h= Planck's constant, R=Rydberg's constant m = mass of photon)

Options

A. `(9Rh)/(16m)`
B. `(11hR)/(16m)`
C. `(13hR)/(16m)`
D. `(15hR)/(16m)`

Answer

D

Explanation

`p=1, n=4, c?`
`(1)/(lamda)=R [(1)/(P^(2))-(1)/(n^(2))]`
`therefore (1)/(lamda)=R [(1)/(1) -(1)/(16)]`
`(1)/(lamda) =(15 R)/(16)" "...(1)`
But `E = mc ^(2)`
`therefore (hc)/(lamda )= mc ^(2)`
`(h)/(lamda)=mc`
`i.e.c=(h)/(m lamda) =h/m xx(15R)/(16)` ...(from equation(1))
`therefore c=(15hR)/(16m)`