Question

The wave number of the spectral line in the emission spectrum of Hydrogen will be equal to $\frac{8}{9}$ times the Rydberg's constant if the electron jumps from:

• A. $n=3$ to $n=1$
• B. $n=10$ to $n=1$
• C. $n=9$ to $n=1$
• D. $n=2$ to $n=1$

Answer 1

Answer: (A)

$n=3$ to $n=1$

Wave number of spectral line in emission spectrum of hydrogen,

$\overline{v}=R_H\left(\begin{array}{c}\frac{1}{n_1^2}-\frac{1}{n_2^2}\end{array}\right)$ ...(i)

Given, $\overline{v}=\frac{8}{9}{R}_H$

On putting the value of $\overline{v}$ in Eq. (i), we get-

$\frac{8}{9}{R}_H=R_H\left(\begin{array}{c}\frac{1}{n_1^2}-\frac{1}{n_2^2}\end{array}\right)$

$\frac{8}{9}=\frac{1}{(1{)}^2}-\frac{1}{n_2^2}$

$\frac{8}{9}-1=\frac{1}{n_2^2}$

$\frac{1}{3}=\frac{1}{n_2}\therefore n_2=3$

Hence, electron jumps from $n_2=3$ to $n_1=1$