Question

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

Answer 1

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 $\left(n\right)$ as:

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

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

$\frac{1}{\lambda }=R_y(\frac{1}{1^2}-\frac{1}{n^2})$

Where, $R_y=$Rydberg constant $=1.097\times {10}^7{m}^{-1}$

$\lambda =$ Wavelength of radiation emitted by the transition of the electron

For $n=3$, we can obtain $\lambda$ as:

$\frac{1}{\lambda }=1.097\times {10}^7(\frac{1}{1^2}-\frac{1}{3^2})=1.097\times {10}^7\times \frac{8}{9}$

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

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

$\frac{1}{\lambda }=1.097\times {10}^7(\frac{1}{2^1}-\frac{1}{2^2})=1.097\times {10}^7\times \frac{3}{4}$

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

If the electron jumps from $n=3$ to $n=2$ then the wavelength of the radiation is given as:

$\frac{1}{\lambda }=1.097\times {10}^7(\frac{1}{2^1}-\frac{1}{3^2})=1.097\times {10}^7\times \frac{5}{46}$

$\Rightarrow \lambda =\frac{36}{5\times 1.097\times {10}^7}=653.33$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.