Question

The ratio of the wavelengths of the last line of Balmer series and the last line of Lyman series is -

Assume same element.

• A. $0.5$
• B. $2$
• C. $1$
• D. $4$

Answer 1

The correct option is D $4$

As per spectral lines, wavelength is given by,

$\frac{1}{\lambda }=R{Z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

For the Balmer series, $n_1=2$ and $n_2=\infty$

$\therefore \frac{1}{{\lambda }_B}=R{Z}^2(\frac{1}{2^2}-\frac{1}{{\infty }^2})=\frac{R{Z}^2}{4}$

Similarly, for the last line of Lyman series, $n_1=1$ and $n_2=\infty$

$\therefore \frac{1}{{\lambda }_L}=R{Z}^2(\frac{1}{1^2}-\frac{1}{{\infty }^2})=R{Z}^2$

Now,

$\frac{\frac{1}{{\lambda }_B}}{\frac{1}{{\lambda }_L}}=\frac{\frac{R{Z}^2}{4}}{R{Z}^2}$

$\Rightarrow \frac{{\lambda }_L}{{\lambda }_B}=\frac{1}{4}$

$\Rightarrow \frac{{\lambda }_B}{{\lambda }_L}=4$

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