Question

What is the lowest energy of the spectral line emitted by the hydrogen atom in the Lyman series?
( $h=$ Planck's constant, $c=$ velocity of light, $R=$ Rydberg's constant).

• A. $\frac{5 \text{h} \text{c} \text{R}}{36}$
• B. $\frac{4 \text{h} \text{c} \text{R}}{3}$
• C. $\frac{3 \text{h} \text{c} \text{R}}{4}$
• D. $\frac{7 \text{h} \text{c} \text{R}}{144}$

Answer 1

Answer: (C)

We know that,
$\Delta \text{E}=\text{h}\text{c}\cdot \text{R}\left[\frac{1}{\text{n}_{1}^{2}} - \frac{1}{\text{n}_{2}^{2}}\right]$
For lowest energy, of the spectral line in Lyman series, $\text{n}_{1}=1, \, \text{n}_{2}=2$
Hence,
$\Delta \text{E}=\text{h}\text{c}.\text{R}\left[\frac{1}{1^{2}} - \frac{1}{2^{2}}\right]$
$\Delta \text{E}=\frac{3 \text{h} \text{c} \text{R}}{4}$