Question

An electron jumps from 5^th orbit to 4^th orbit of hydrogen atom. Taking the Rydberg constant as $10^{7}$ per metre. What will be the frequency of radiation emitted.

• A. $6.75 \times 10^{12} ⥂ Hz$
• B. $6.75 \times 10^{14}Hz$
• C. $6.75 \times 10^{13}Hz$
• D. None of these

Answer 1

Answer: (C)

$6.75 \times 10^{13}Hz$

Answer 2

By using $\nu = RC\left\lbrack \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right\rbrack$

$\Rightarrow \nu = 10^{7} \times (3 \times 10^{8})\left\lbrack \frac{1}{4^{2}} - \frac{1}{5^{2}} \right\rbrack$= 6.75 × 10¹³ Hz.