JEE MAIN - Physics (2023 - 15th April Morning Shift - No. 22)
As per given figure $A, B$ and $C$ are the first, second and third excited energy levels of hydrogen atom respectively. If the ratio of the two wavelengths $\left(\right.$ i.e. $\left.\frac{\lambda_{1}}{\lambda_{2}}\right)$ is $\frac{7}{4 n}$, then the value of $n$ will be __________.
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Answer
5
Explanation
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$$ \begin{aligned} & \text { As } \frac{1}{\lambda}=\mathrm{RZ}^2\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right] \\\\ & \frac{1}{\lambda_1}=\mathrm{R}(1)^2\left[\frac{1}{(2)^2}-\frac{1}{(3)^2}\right]=\mathrm{R}\left(\frac{5}{36}\right)~~......(i) \\\\ & \& \frac{1}{\lambda_2}=\mathrm{R}(1)^2\left[\frac{1}{(3)^2}-\frac{1}{(4)^2}\right]=\mathrm{R}\left(\frac{7}{144}\right) ~~......(ii) \end{aligned} $$
(ii) $\div$ (i) gives
$$ \begin{aligned} & \frac{\lambda_1}{\lambda_2}=\frac{7 / 144}{5 / 36}=\frac{7}{20}=\frac{7}{4 \times 5} \\\\ & \therefore \mathrm{n}=5 \end{aligned} $$
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