Physics - Modern Physics Question with Solution | TestHub

For ground state$n_1 = 1 \mathrm{to} n_2 = 4$

Energy absorbed by photon,$E\mathit{ }\mathit{=}\mathit{ }{E}_2\mathit{ }\mathit{-}\mathit{ }{E}_1$

$=+\text{13.6}\left(\frac{1}{{\text{n}}_1^2}-\frac{1}{{\text{n}}_2^2}\right)\times \text{1.6}\times {\text{10}}^{-\text{19}}\text{ J}$

$=\text{13.6}\left(\frac{1}{1}-\frac{1}{4^2}\right)\times \text{1.6}\times {\text{10}}^{-\text{19}}$

$=\text{13.6}\times \text{1.6}\times {\text{10}}^{-\text{19}}\left(\frac{\text{15}}{\text{16}}\right)$

$= 2.4 \times {10}^{-19}$

or$E = hv = 20.4 \mathrm{x} {10}^{-19}$$\mathrm{J}$

$\text{Frequency }\mathit{\text{v =}}\frac{\text{20.4}\times {\text{10}}^{-\text{19}}}{\text{h}}$

$=\frac{\text{20.4}\times {\text{10}}^{-\text{19}}}{\text{6.63}\times {\text{10}}^{-\text{34}}}$

$= 3.076 \times {10}^{15}$

$= 3.1 \times {10}^{15} \mathrm{Hz}$