Question

What is the lowest energy of the spectral line emitted by the hydrogen atom in the Lyman series?

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

Answer 1

Answer: (C)

We know that

$\Delta E=hc{R}_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

For the lowest energy of spectral line in the lyman series

$n_1=1,n_2=2$

Hence, $\Delta E=hc{R}_H\left(\frac{1}{(1{)}^2}-\frac{1}{(2{)}^2}\right)$

$=hc{R}_H\left(\frac{1}{1}-\frac{1}{4}\right)$

$=\frac{3hc{R}_H}{4}$