Q. An electron jumps from the 4th orbit to 2nd orbit of hydrogen atom. Given the Rydberg’s constant $R = 10^{5}$ $cm\ ^{- 1}$, the frequency in hertz of the emitted radiation will be
A) $\frac{3}{16} \times 10^{5}$
B) $\frac{3}{16} \times 10^{15}$
C) $\frac{9}{16} \times 10^{15}$
D) $\frac{3}{4} \times 10^{15}$
Explanation
(c) $v = \frac{c}{\lambda} = c.R\left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right)$
$= 3 \times 10^{8} \times 10^{7}\left( \frac{1}{2^{2}} - \frac{1}{4^{2}} \right) = \frac{9}{16} \times 10^{15}Hz$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.