Question

An electron from various states of hydrogen atom emit radiation to come to the ground state. Let ${\lambda }_n,{\lambda }_g$ be the de Broglie wavelength of the ground state repectively. Let ${\Lambda }_n$ 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)

• A. ${\Lambda }_n^2\approx A+B{\lambda }_n^2$
• B. ${\Lambda }_n^2\approx \lambda$
• C. ${\Lambda }_n\approx A+\frac{B}{{\lambda }_n^2}$
• D. ${\Lambda }_n\approx A+B{\lambda }_n$

Answer 1

The correct option is C ${\Lambda }_n\approx A+\frac{B}{{\lambda }_n^2}$

The wavelength of emitted photon from $nth$ state of the ground state,

$\frac{1}{{\lambda }_n}=R.Z^2(\frac{1}{1^2}-\frac{1}{n^2})$

${\lambda }_n=\frac{1}{{RZ}^2}(1-\frac{1}{n^2})-1$

Since, $n$ is very large, using the binomial theorem,

${\Lambda }_n=\frac{1}{{RZ}^2}(1+\frac{1}{n^2})$

${\Lambda }_n=\frac{1}{{RZ}^2}+\frac{1}{{RZ}^2}\left(\frac{1}{n^2}\right)$

As we know, ${\lambda }_n=\frac{2nr}{n}=2\pi \left(\frac{n^2{h}^2}{{4\pi }^2{mZe}^2}\right)\frac{1}{n}\alpha n$

${\lambda }_n\approx A+\frac{B}{{\lambda }_n^2}$