Question

A $12.5 eV$ 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 $12.5eV$. Also, the energy of the gaseous hydrogen in its ground state at room temperature is $-13.6eV$.
When gaseous hydrogen is bombarded with an electron beam, the energy of gaseous hydrogen becomes $-13.6+12.5eV$ i.e.,$-1.1eV$. Orbital energy is related to orbit level (n) as:
$E=\frac{-13.6}{(n)^2}eV$
For n=3, $E=\frac{-13.6}{9}=-1.5eV$
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:
$\frac{1}{λ}=R_y(\frac{1}{1^2}-\frac{1}{n^2})$
Where,
$R_y$=Rydberg constant$=1.097×10^7m^{-1}λ=$Wavelength of radiation emitted by the transition of the electron
For n=3,we can obtain $λ$ as:
$\frac{1}{λ}=1.097×10^7(\frac{1}{1^2}-\frac{1}{3^2})$
$=1.097×10^7(1-\frac{1}{9})=1.097×10^7×\frac{8}{9}$
$λ=\frac{9}{8×1.097×10^7}=102.55nm$
If the electron jumps from n=2 to n=1, then the wavelength of the radiation is given as:
$\frac{1}{λ}=1.097×10^7(\frac{1}{1^2}-\frac{1}{2^2})$
$=1.097×10^7(1-\frac{1}{4})=1.097×10^7×\frac{3}{4}$
$λ=\frac{4}{1.097×10^7×3}=121.54nm$
If the transition takes place from n=3 to n=2,then the wavelength of the radiation is given as:
$\frac{1}{λ}=1.097×10^7(\frac{1}{2^2}-\frac{1}{3^2})$
$=1.097×10^7(\frac{1}{4}-\frac{1}{9})=1.097×10^7×\frac{5}{36}$
$λ=\frac{36}{5×1.097×10^7}=656.33nm$
This radiation corresponds to the Balmer series of the hydrogen spectrum.
Hence,in Lyman series,two wavelengths i.e.,102.5nm and 121.5nm are emitted.
And in the Balmer series,one wavelength i.e.,656.33nm is emitted.