Question

An electron jumps from the 4^th orbit to the 2^nd orbit of hydrogen atom. Given the Rydberg's constant $R = 10^{5}cm^{- 1}$. The frequency in Hz 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)

$\frac{9}{16} \times 10^{15}$

Answer 2

$\frac{1}{\lambda} = R\left( \frac{1}{2^{2}} - \frac{1}{4^{2}} \right) = \frac{3R}{16} \Rightarrow \lambda = \frac{16}{3R} = \frac{16}{3} \times 10^{- 5}cm$

Frequency $n = \frac{c}{\lambda} = \frac{3 \times 10^{10}}{\frac{16}{3} \times 10^{- 5}} = \frac{9}{16} \times 10^{15}Hz$