Physics - Modern Physics Question with Solution | TestHub

For Hydrogen like atom, energy emitted is $E=-13.6Z^2\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right)$

$\Rightarrow E=C\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$

$\Rightarrow h\nu =C\left[\frac{1}{n_f^2}-\frac{1}{n_i^2}\right]$

Therefore, the  ratio$\frac{v_1}{v_2}=\frac{{\left[\frac{1}{n_f^2}-\frac{1}{n_i^2}\right]}_{2-1}}{{\left[\frac{1}{n_f^2}-\frac{1}{n_i^2}\right]}_{3-1}}$

$=\frac{\left[\frac{1}{1}-\frac{1}{4}\right]}{\left[\frac{1}{1}-\frac{1}{9}\right]}=\frac{{\displaystyle \frac{3}{4}}}{{\displaystyle \frac{8}{9}}}$

$=\frac{3}{4}\times \frac{9}{8}$

$\Rightarrow \frac{v_1}{v_2}=\frac{27}{32}$

$\Rightarrow v_2=\frac{32}{27}{v}_1=\frac{32}{27}\times 3\times {10}^{15} \mathrm{Hz}=\frac{32}{9}\times {10}^{15} \mathrm{Hz}$

Hence, $x=32$.