Question

When an electron makes a transition from $(n+1)$ state to $n^{th}$ state, the frequency of emitted radiations is related to $n$ according to $(n>>1)$:

• A. $v=\frac{2cR{Z}^2}{n^3}$
• B. $v=\frac{cR{Z}^2}{n^4}$
• C. $v=\frac{cR{Z}^2}{n^2}$
• D. $v=\frac{2cR{Z}^2}{n^2}$

Answer 1

The correct option is A $v=\frac{2cR{Z}^2}{n^3}$

$\frac{1}{\lambda }=R{Z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right),$ where $n_1=n,n_2=n+1$
$\therefore \frac{1}{\lambda }=R{Z}^2\left(\frac{1}{n^2}-\frac{1}{(n+1{)}^2}\right)$
$\Rightarrow \frac{1}{\lambda }=\left(\frac{2n+1}{n^2(n+1{)}^2}\right)R{Z}^2$
Since, $n>>1$;
Therefore, $2n+1\approx 2n$;
and $(n+1{)}^2\approx n^2$
$\therefore \frac{1}{\lambda }=R{Z}^2\left(\frac{2n}{n^2.n^2}\right)$
$\Rightarrow \frac{v}{c}=\frac{2R{Z}^2}{n^3}$ or $v=\frac{2cR{Z}^2}{n^3}$

Hence, the correct option is $A$