Q. 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) $\frac{5hcR}{36}$
B) $\frac{4hcR}{3}$
C) $\frac{3hcR}{4}$
D) $\frac{7hcR}{144}$
Explanation
(c)
We know that,
$\mathrm{\Delta}E = hc.R\left\lbrack \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right\rbrack$
For lowest energy, of the spectral line in Lyman series, $n_{1} = 1,\ n_{2} = 2$
Hence,
$\mathrm{\Delta}E = hc.R\left\lbrack \frac{1}{1^{2}} - \frac{1}{2^{2}} \right\rbrack$
$\mathrm{\Delta}E = \frac{3hcR}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.