Question

What is the ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lyman series ?

• A. 4 : 1
• B. 4 : 9
• C. 4 : 3
• D. 5 : 9

Answer 1

The correct option is A 4 : 1

For transition $n_2\to n_1$

the wavelength is given by $\frac{1}{\lambda }=R{Z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

For shortest wavelength in Lyman series put $n_1=1$ and $n_2\to \infty$

and for shortest wavelength in Balmer series put $n_1=2$ and $n_2\to \infty$

we get $\frac{1}{{\lambda }_L}=R{Z}^2$

and $\frac{1}{{\lambda }_B}=R{Z}^2\left(\frac{1}{4}\right)$

ratio will be $\frac{{\lambda }_B}{{\lambda }_L}=\frac{4}{1}$

Option A is correct.