Question

An electron jumps from the $4^{th}$ orbit to $2^{nd}$ 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} $

Answer 1

Answer: (C)

Frequency, $v =\frac{c}{\lambda}=c \cdot R\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)$
$=3 \times 10^{8} \times 10^{7}\left( \frac{1}{2^{2}}-\frac{1}{4^{2}}\right)$
$=\frac{9}{16} \times 10^{15} Hz$