# A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature.

Question:

A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

Solution:

It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is 12.5 eV. Also, the energy of the gaseous hydrogen in its ground state at room temperature is −13.6 eV.

When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogen becomes −13.6 + 12.5 eV i.e., −1.1 eV.

Orbital energy is related to orbit level (n) as:

$E=\frac{-13.6}{(n)^{2}} \mathrm{eV}$

For $n=3, E=\frac{-13.6}{9}=-1.5 \mathrm{eV}$

This energy is approximately equal to the energy of gaseous hydrogen. It can be concluded that the electron has jumped from n = 1 to n = 3 level.

During its de-excitation, the electrons can jump from n = 3 to n = 1 directly, which forms a line of the Lyman series of the hydrogen spectrum.

We have the relation for wave number for Lyman series as:

$\frac{1}{\lambda}=R_{y}\left(\frac{1}{1^{2}}-\frac{1}{n^{2}}\right)$

Where,

Ry = Rydberg constant = 1.097 × 107 m−1

λ= Wavelength of radiation emitted by the transition of the electron

For n = 3, we can obtain λas:

$\frac{1}{\lambda}=1.097 \times 10^{7}\left(\frac{1}{1^{2}}-\frac{1}{3^{2}}\right)$

=1.097 \times 10^{7}\left(1-\frac{1}{9}\right)=1.097 \times 10^{7} \times \frac{8}{9}

$\lambda=\frac{9}{8 \times 1.097 \times 10^{7}}=102.55 \mathrm{~nm}$

If the electron jumps from n = 2 to n = 1, then the wavelength of the radiation is given as:

$\frac{1}{\lambda}=1.097 \times 10^{7}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)$

$=1.097 \times 10^{7}\left(1-\frac{1}{4}\right)=1.097 \times 10^{7} \times \frac{3}{4}$

$\lambda=\frac{4}{1.097 \times 10^{7} \times 3}=121.54 \mathrm{~nm}$

If the transition takes place from n = 3 to n = 2, then the wavelength of the radiation is given as:

$\frac{1}{\lambda}=1.097 \times 10^{7}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$

$=1.097 \times 10^{7}\left(\frac{1}{4}-\frac{1}{9}\right)=1.097 \times 10^{7} \times \frac{5}{36}$

$\lambda=\frac{36}{5 \times 1.097 \times 10^{7}}=656.33 \mathrm{~nm}$

This radiation corresponds to the Balmer series of the hydrogen spectrum.

Hence, in Lyman series, two wavelengths i.e., 102.5 nm and 121.5 nm are emitted. And in the Balmer series, one wavelength i.e., 656.33 nm is emitted.