Question

What is the ratio of wavelength of radiations emitted when an electron in a hydrogen atom jumps from fourth orbit to second orbit and from third orbit to second orbit?

• A. $27:25$
• B. $20:27$
• C. $20:25$
• D. $25:27$

Answer 1

The correct option is B $20:27$

We know that,
$\Rightarrow \frac{1}{\lambda }=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

When the electron jumps from fourth to second orbit, $n_1=2,n_2=4$

$\Rightarrow \frac{1}{{\lambda }_{4\to 2}}=R(\frac{1}{2^2}-\frac{1}{4^2})$

$\Rightarrow \frac{1}{{\lambda }_{4\to 2}}=R(\frac{1}{2^2}-\frac{1}{4^2})$
$\Rightarrow \frac{1}{{\lambda }_{4\to 2}}=R(\frac{1}{4}-\frac{1}{16})$
$\Rightarrow {\lambda }_{4\to 2}=\left(\frac{16}{3R}\right)----\left(1\right)$

Similarly, when the electron jumps from third orbit to second orbit, $n_1=2,n_2=3$

$\Rightarrow \frac{1}{{\lambda }_{3\to 2}}=R(\frac{1}{2^2}-\frac{1}{3^2})$

$\Rightarrow \frac{1}{{\lambda }_{3\to 2}}=R(\frac{1}{4}-\frac{1}{9})$

$\Rightarrow {\lambda }_{3\to 2}=\left(\frac{36}{5R}\right)----\left(2\right)$

Dividing equation $\left(1\right)$ by equation $\left(2\right)$,

$\Rightarrow \frac{{\lambda }_{4\to 2}}{{\lambda }_{3\to 2}}=\frac{16}{3R}\times \frac{5R}{36}=\frac{20}{27}$

Hence, $\left(B\right)$ is the correct answer.