An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let `lambda_(n),lambda_(g)` be the de Broglie wavelength of the electron in the `n^(th)` state and the ground state respectively. Let `^^^_(n)` be the wavelength of the emitted photon in the transition from the `n^(th)` state to the ground state. For large n, (A, B are constants)

Options

A. `Lambda_n^2 = A+Blambda_n^2`
B. `Lambda_n^2approx lambda`
C. `Lambda_n approx A+B/lambda_n^2`
D. `Lambda_n approx A+B lambda_n`

Answer

C

Explanation

`1/Lambda_n=RZ^2=(1/1^2-1/n^2)`
`Lambda_n=1/(RZ^2)(1-1/n^2)^(-1)`
For n is large so we can use binomial approximation.
`Lambda_n=1/(RZ^2)(1+1/n^2)`
`rArr Lambda_n=1/(RZ^2)+1/(RZ^2)(1/n^2)`….(i)
De-Broglie wavelength of electron in `n^(th)` orbit can be calculated as follows :
`lambda_m=(2pir)/n=2pi((n^2h^2)/(4pi^2mZe^2))1/n rArr lambda_n prop n^2`
Using the above result in equation (i) we can rewrite the equation (i) as follows :
`Lambda_n approx A+B/lambda_n^2`