When an electron makes a transition from (n+1) state to n^ th sta…

MCQ

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 $(n > > 1)$

✓
$v = \frac{{2cR{Z^2}}}{{{n^3}}}$
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

Correct option: A.

$v = \frac{{2cR{Z^2}}}{{{n^3}}}$

a
$\frac{1}{\lambda}=\mathrm{R} Z^{2}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right),$ where $\mathrm{n}_{1}=\mathrm{n}$

$\mathrm{n}_{2}=\mathrm{n}+1$

$\therefore \quad \frac{1}{\lambda}=\mathrm{R} Z^{2}\left(\frac{1}{\mathrm{n}^{2}}-\frac{1}{(\mathrm{n}+1)^{2}}\right)$

$\Rightarrow \quad \frac{1}{\lambda}=\left(\frac{2 n+1}{n^{2}(n+1)^{2}}\right) R Z^{2}$

since, $n>>1;$

Therefore, $\quad 2 \mathrm{n}+1 \approx 2 \mathrm{n}$

and $(n+1)^{2} \approx n^{2}$

$\therefore$ $\frac{1}{\lambda}=\mathrm{R} Z^{2}\left(\frac{2 \mathrm{n}}{\mathrm{n}^{2} \cdot \mathrm{n}^{2}}\right)$

$\Rightarrow \frac{\mathrm{v}}{\mathrm{c}}=\frac{2 \mathrm{R} \mathrm{Z}^{2}}{\mathrm{n}^{3}}$ or $\mathrm{v}=\frac{2 \mathrm{cR} Z^{2}}{\mathrm{n}^{3}}$