A free hydrogen atom after absorbing a photon of wavelength `lambda_(a)` gets excited from state n = 1 to n = 4. Immediately after electron jumps to n = m state by emitting a photon of wavelength `lambda_(e )`. Let change in momentum of atom due to the absorption and the emission are `Delta P_(a)` and `Delta P_(e)` respectively. If `lambda_(a)//lambda_(e )= 1//5`. Which of the following is correct

Options

A. m = 2
B. `Delta P_(a)//Delta P_(e )= 1//2`
C. `lambda_(e )=418 nm`
D. Ratio of K.E. of electron in the state n = m to n= 1 is `1//4`

Answer

A, D

Explanation

`(lambda_(a))/(lambda_(e ))=(E_(4)-E_(1))/(E_(4)-E_(m))=(1-(1)/(16))/((1)/(m^(2))-(1)/(16))=(1)/(5)`
on solving
m = 2
`lambda_(e ) = (12400xx4)/(13.6)=3647`
`(K_(2))/(K_(1))=(1^(2))/(2^(2))=(1)/(4)` As kinetic energy is proportional to `(1)/(n^(2))`