# 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? - Physics

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.

#### Solution

Let the hydrogen atoms be excited to nth 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

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

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

$\Rightarrow \lambda = 4861 A^\circ$

Concept: Energy Levels
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