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

Answer 1

Answer: (B)

Wave number of spectral line in emission spectrum of hydrogen,

$\bar{v}=R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$...(i)

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

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

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

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

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

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

$\therefore n_2=3$

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