Question

In hydrogen like atom electron makes transition from an energy level with quantum number $n$ to another with quantum number $(n-1)$. If $n>>1$, the frequency of radiation emitted is proportional to :

• A. $\frac{1}{n^2}$
• B. $\frac{1}{n^{3/2}}$
• C. $\frac{1}{n^3}$
• D. $\frac{1}{n^{}}$

Answer 1

The correct option is C $\frac{1}{n^3}$

$\begin{array}{c}E=-13.6(\frac{1}{n_1^2}-\frac{1}{n_2^2})=\frac{hc}{\lambda e}\\ -13.6(\frac{1}{n^2}-\frac{1}{(n-1{)}^2})=\frac{hv}{e}\\ x-\frac{13\cdot 6}{e}\left(\frac{n^2-n^2-1+2n}{n^2(n-1{)}^2}\right)=hv\\ \frac{2n-1}{n^2(n-1{)}^2}\times v\end{array}$

$V=\frac{2n-1}{n^2(n^2+1-2n)}$

$v_i\times \frac{2n-1}{n^4+n^2-2n^3}\dots .$

$\therefore n>>1$

$v\times \frac{2n}{n^4}\times \frac{2}{n^3}\times \frac{1}{n^3}$

$v\times \frac{1}{n^3}$

Correct C