Question

An electron in hydrogen atom first jumps form second excited state to first excited state and then form first excited state to ground state. Let the ratio of wavelength, momentum and energy of photons emitted in these two cases be a, b and c respectively, Then
A. `a=9/4`
B. `b=5/27`
C. `c=5/27`
D. `c=1/a`

Answer 1

Correct Answer - B::C::D
First transition is form n=3 to n=2. Second transition is form n=2 to n=1.
`:. (E_(1))/(E_(2))=c=(1//2^(2)-1//3^(2))/(1//1^(2)-1//2^(2))=(5//36)/(3//4) =5/36xx4/3=5/27`
As `p=E/c`, therefore, `(p_(1))/(p_(2))=b=(E_(1))/(E_(2))=c`
i.e,. `B=c=5/27`
As `E=(hc)/lambda :. lambda prop 1/E`
or `a=(lambda_(1))/(lambda_(2))=(E_(2))/(E_(1))=27/5=1/c or c=1/a`