Question

The wavelength of light for the least energetic photon emitted in the Lyman series of the hydrogen atom spectrum is __________ nm

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Answer 1

Least energy is the cue my friend.

Energy is $\frac{hc}{\lambda }$

Least energy means $\frac{hc}{\lambda }$ should be less. hc is constant

So $\frac{1}{\lambda }$ should be less

By Rydberg’s formula; $\frac{1}{\lambda }=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

For Lyman, $n_1=1$

$\to \frac{1}{\lambda }=R(\frac{1}{1}-\frac{1}{n_2^2})$

$n_2$ can be from 2, …. $\infty$

To minimize $\frac{1}{\lambda }$ what should be $n_2$

2 right and not $\alpha$

So Let’s find $\lambda$

$\frac{1}{\lambda }=R\left(\frac{3}{4}\right)\lambda =\frac{4}{3R}=\frac{4}{3\times 1.09\times {10}^7}=1.22\times {10}^{-7}=122nm$