JEE MAIN - Physics (2024 - 31st January Morning Shift - No. 12)
If the wavelength of the first member of Lyman series of hydrogen is $$\lambda$$. The wavelength of the second member will be
$$\frac{27}{5} \lambda$$
$$\frac{5}{27} \lambda$$
$$\frac{27}{32} \lambda$$
$$\frac{32}{27} \lambda$$
Explanation
$$\begin{aligned} & \frac{1}{\lambda}=\frac{13.6 \mathrm{z}^2}{\mathrm{hc}}\left[\frac{1}{1^2}-\frac{1}{2^2}\right] ...... \text{(i)}\\ & \frac{1}{\lambda^{\prime}}=\frac{13.6 \mathrm{z}^2}{\mathrm{hc}}\left[\frac{1}{1^2}-\frac{1}{3^2}\right] ...... \text{(ii)} \end{aligned}$$
On dividing (i) & (ii)
$$\lambda^{\prime}=\frac{27}{32} \lambda$$
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