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
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:
This energy is approximately equal to the energy of
gaseous hydrogen. lt can be concluded that the electron has
jumped from n= 1 to n = 3 level.
Step 2: Find the wavelength of the spectral line for n= 3 to n = 1.
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:
where,
= Rydberg constant =
= Wavelength of radiation emitted by the transition of the electron.
For n = 3, we can obtain as:
Step 3: Find the wavelength of the spectral line for n= 3 to n = 2 and n = 2 to n = 1.
If the electron jumps from n=3 to n=2, then the wavelength of the radiation is given as:
\(\frac{1}{\lambda}=1.097 \times 10^{7} [\frac{1}{2^2}-\frac{1}{3^{2}}]=1.097 \times 10^{7} [\frac{1}{4}-\frac{1}{9}]=1.097 \times 10^{7} \times \frac{5}{36}\\ \lambda=\frac{36}{5 \times 1.097 \times 10^{7}}=656.33 \mathrm{~nm}\)
Then, the electron jumps from n=2 to n=1, then the wavelength of the radiation is given as:
\(\frac{1}{\lambda}=1.097 \times 10^{7} [\frac{1}{1^2}-\frac{1}{2^{2}}]=1.097 \times 10^{7} [\frac{1}{1}-\frac{1}{4}]=1.097 \times 10^{7} \times \frac{3}{4}\\ \lambda=\frac{4}{3 \times 1.097 \times 10^{7}}=121.54 \mathrm{~nm}\)
This radiation corresponds to the Balmer series of the hydrogen spectrum.
Hence, in the Lyman series, two wavelengths i.e., 102.5 nm and 121.54 nm are
emitted. And in the Balmer series, one wavelength i.e., 656.33 nm is emitted.
