Question

The difference in the wavelength of the second line of Lyman series and last line of bracket series in a hydrogen sample is :

• A. $\frac{119}{8R}$
• B. $\frac{1271}{8R}$
• C. $\frac{219}{8R}$
• D. None of these

Answer 1

The correct option is A $\frac{119}{8R}$

$\lambda =\frac{1}{\nu }=\frac{1}{R{Z}^2}\frac{1}{(\frac{1}{n_1^2}-\frac{1}{n_2^2})}$
For the second line of Lyman series of H atom $3\to 1$,
$\lambda =\frac{1}{R}\frac{1}{\frac{1}{3^2}-\frac{1}{1^2}}$
$\lambda =\frac{9}{8R}$
For the last line of Balmer series, $\infty \to 4$,

${\lambda }^{\prime }=\frac{1}{R}\frac{1}{\frac{1}{4^2}-\frac{1}{{\infty }^2}}$
${\lambda }^{\prime }=\frac{16}{R}$
Hence, the difference in the wavelength is $\frac{9}{8R}-\frac{16}{R}=\frac{119}{8R}$.