Question

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

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

Answer 1

Answer: (A)

4 : 1

Answer 2

For a Balmer series

$\frac{1}{\lambda_{B}} = R\left\lbrack \frac{1}{2^{2}} - \frac{1}{n^{2}} \right\rbrack$ …… (i)

Where n = 3, 4 …….

By putting n = $\infty$ in equation (i) we obtain the series limit of the Balmer series this is the shortest wavelength of the Balmer series

$or\lambda_{B} = \frac{4}{R}$ …… (ii)

For Lyman series

$\frac{1}{\lambda_{L}} = R\left\lbrack \frac{1}{1^{2}} - \frac{1}{n^{2}} \right\rbrack$ ……. (iii)

Where n = 2, 3, 4………

By putting $n = \infty$g in equation (iii) We obtain the series limit of the Lyman series. This is the shortest wavelength of the Lyman series.

Or $\lambda_{L} = \frac{1}{R}$ ……..(iv)

Dividing (ii) by (iv) , we get

$\frac{\lambda_{B}}{\lambda_{L}} = \frac{4}{1}$