Question

Find out the wavelength of the next line in the series having lines of spectrum of $H$-atom of wavelengths $656.46,486.27,434.17$ and $410.29nm$.

• A. $397nm$
• B. $122nm$
• C. $1282nm$
• D. $302nm$

Answer 1

The correct option is A $397nm$

The given series belongs to Balmer series since given $\lambda$ are in the visible region.

Therefore, $n_1=2$

If $\lambda =410.29\times {10}^{-7}cm$ and $n_1=2$ the by using

$\frac{1}{\lambda }=R_H[\frac{1}{2^2}-\frac{1}{n_2^2}]$

$\frac{1}{410.29\times {10}^{-7}}=109678[\frac{1}{2^2}-\frac{1}{n_2^2}]$

$\therefore n_2=6$

Therefore, next line will be from $n_2=7$ to $n_1=2$

$\therefore \frac{1}{\lambda }=R_H[\frac{1}{2^2}-\frac{1}{7^2}]$

$=109678[\frac{1}{2^2}-\frac{1}{7^2}]$

$\therefore \lambda =397.2\times {10}^{-7}cm=397.2nm$