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

Hint : First find out the energy associated with the excited electron, then the final excited state. After this count the total number of transitions that are possible between the ground level and the excited state and then find the wavelengths associated with each transition.

Complete step by step solution :
-Energy of the electron in the nth orbit is given as:
${E_n} = \dfrac{{ - 13.6}}{{{n^2}}}eV$ (1)
Where n = energy level or orbit number
-Wavelength associated with the electron transition in a H atom:
$\dfrac{1}{\lambda } = {R_h}\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right)$ (2)
Where, ${R_y}$ = Rydberg constant = $1.097 \times {10^7}{m^{ - 1}}$
$\lambda$ = wavelength of the radiation emitted by the transition of the electron

-We all know that the energy of the gaseous hydrogen atom in its ground state = (-13.6 eV)
-The energy of the electron beam used to bombard gaseous hydrogen = 12.5 eV
-After bombardment the energy of the gaseous hydrogen atoms becomes = -13.6 + 12.5
= -1.1 eV
-Now since we have the energy of the excited state, we can find out the orbital number of the excited state using equation (1):
${E_n} = \dfrac{{ - 13.6}}{{{n^2}}}eV$
$- 1.1 = \dfrac{{ - 13.6}}{{{n^2}}}$
${n^2} = \dfrac{{ - 13.6}}{{ - 1.1}}$ = 12.36
n = $\sqrt {12.36}$ = 3.5
So, we take the value of n to be 3.
-From this we can say that the electron bean excited the electron in H atom from n = 1 to n = 3. During de-excitation the number of transitions possible is:
- From n = 3 to n = 1 (Lyman series)
- From n = 2 to n = 1 (Lyman series)
- From n = 3 to n = 2 (Balmer series)
We will now find out the wavelengths associated with each transition.

-When the electron undergoes transition from n = 3 to n = 1: ${n_1}$= 1 and ${n_2}$= 3; we can find the associated wavelength using equation (2):
$\dfrac{1}{\lambda } = {R_h}\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right)$
$\dfrac{1}{\lambda } = 1.097 \times {10^7}\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{3^2}}}} \right)$
$\lambda = \dfrac{9}{{8 \times 1.097 \times {{10}^7}}}$
= $1.0255 \times {10^{ - 7}}$
= 102.55 nm
This radiation of 102.55 nm lies in the Lyman series.

-When the electron undergoes transition from n= 2 to n = 1: ${n_1}$= 1 and ${n_2}$= 2; we can find the associated wavelength using equation (2):
$\dfrac{1}{\lambda } = 1.097 \times {10^7}\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}} \right)$
$\lambda = \dfrac{4}{{3 \times 1.097 \times {{10}^7}}}$
= $1.2154 \times {10^{ - 7}}$
= 121.54 nm
This radiation of 121.54 nm also lies in the Lyman series.

-When the transition occurs from n = 3 to n = 2: ${n_1}$= 2 and ${n_2}$= 3; we will again use equation (2) and find out the wavelength associated:
$\dfrac{1}{\lambda } = 1.097 \times {10^7}\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}} \right)$
$\lambda = \dfrac{{36}}{{5 \times 1.097 \times {{10}^7}}}$
= $6.5633 \times {10^{ - 7}}$
= 656.33 nm
This radiation of 656.33 lies in the Balmer series.

So, 2 wavelengths 121.54 nm and 102.55 nm are emitted in the Lyman series and 1 wavelength of 656.33 nm is emitted in the Balmer series.

Note : The energy of the electron beam (12.5 eV) is not the energy associated with the excited electron. The electron at its ground state in H atom has is (-13.6 eV) and 12.5 eV is the energy required to excite the electron to the excited state. To calculate the energy of the excited state of H atom we need to add both these energies.