Question

A 12.3 eV electron beam is used to bombard gaseous hydrogen at room temperature. Upto which energy level the hydrogen atoms would be excited?
Calculate the wavelengths of the second member of Lyman series and second member of Balmer series.

Answer 1

Let the hydrogen atoms be excited to n^th energy level.

\[12 . 3 = 13 . 6\left( \frac{1}{1^2} - \frac{1}{n^2} \right)\]

\[ \Rightarrow 12 . 3 = 13 . 6 - \frac{13 . 6}{n^2}\]

\[ \Rightarrow \frac{13 . 6}{n^2} = 13 . 6 - 12 . 3 = 1 . 3\]

\[ \Rightarrow n^2 = \frac{13 . 6}{1 . 3}\]

\[ \Rightarrow n \approx 3\]

 The formula for calculating the wavelength of Lyman series is given below:

\[\frac{1}{\lambda} = R\left( 1 - \frac{1}{n^2} \right)\]

For the second member of Lyman series:
n = 3

\[\therefore \frac{1}{\lambda} = R\left( 1 - \frac{1}{3^2} \right)\]

\[ \Rightarrow \frac{1}{\lambda} = \left( 1 . 09737 \times {10}^7 \right)\left( \frac{8}{9} \right)\]

\[ \Rightarrow \lambda = 1025 A^\circ\]

The formula for calculating the wavelength of Balmer series is given below:

\[\frac{1}{\lambda} = R\left( \frac{1}{4} - \frac{1}{n^2} \right)\]

For second member of Balmer series:
n = 4

\[\therefore \frac{1}{\lambda} = R\left( \frac{1}{4} - \frac{1}{4^2} \right)\]

\[ \Rightarrow \frac{1}{\lambda} = \left( 1 . 09737 \times {10}^7 \right)\left( \frac{3}{16} \right)\]

\[ \Rightarrow \lambda = 4861 A^\circ\]