A 12.9 eV beam of electronics is used to bombard gaseous hydrogen at room temperature.
Up to which energy level the hydrogen atoms would be excited? Calculate the wavelength of the first member of the Paschen series and first member of Balmer series.
Answer
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Hint: Calculate the energy levels which will have energy differences equal to the energy of the electronic beam. Use Rydberg’s formula to calculate the required wavelength for Balmer and Paschen series, taking ni = 2 for balmer and ni = 3 for Paschen.
Formula used:
Rydberg’s formula for spectrum of hydrogen:
\[\dfrac{1}{\lambda } = R(\dfrac{1}{{{n_i}^2}} - \dfrac{1}{{{n_f}^2}})\] , where R is the Rydberg’s constant=$1.097 \times {10^7}$
Energy of electron in nth orbit = $ - \dfrac{{13.6{z^2}}}{{{n^2}}}$ . For hydrogen z=1.
Complete step-by-step solution -
The energy difference between any two levels of the atom should be less than or equal to the energy of the electronic beam = 12.9 eV.
This can be formulated as $ - \dfrac{{13.6}}{{{n_f}^2}} + \dfrac{{13.6}}{{{n_i}^2}} \leqslant 12.9$ ni = 1.
$
- \dfrac{{13.6}}{{{n_f}^2}} + 13.6 \leqslant 12.9 \\
13.6 - 12.9 \leqslant \dfrac{{13.6}}{{{n_f}^2}} \\
{n_f}^2 \leqslant 19.43 \\
{n_f} \leqslant 4.4 \\
{n_f} = 4 \\
$
Therefore, the H-atom excites up to energy level 4.
For calculating the wavelength using the Rydberg’s formula: \[\dfrac{1}{\lambda } = R(\dfrac{1}{{{n_i}^2}} - \dfrac{1}{{{n_f}^2}})\]
For first wavelength of Balmer series- ni = 2 and nf = 3
$
\dfrac{1}{\lambda } = 1.097 \times {10^7}[\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}] \\
\lambda = 6563\mathop {\text{A}}\limits^{\text{o}} \\
$
For first wavelength of Paschen series- ni = 3 and nf = 4
$
\dfrac{1}{\lambda } = 1.097 \times {10^7}[\dfrac{1}{{{3^2}}} - \dfrac{1}{{{4^2}}}] \\
\lambda = 18761\mathop {\text{A}}\limits^{\text{o}} \\
$
The hydrogen atom is excited to the 4th energy level.
The first wavelength of the Balmer series is 6563 angstroms.
The first wavelength of Paschen series is 18761 angstroms.
Note-The energy level excitation could also be determined by finding the difference between individual pairs of energy levels and then comparing them with the energy of the electronic beam. It is important to remember the energy level formula and Rydberg's formula.
Formula used:
Rydberg’s formula for spectrum of hydrogen:
\[\dfrac{1}{\lambda } = R(\dfrac{1}{{{n_i}^2}} - \dfrac{1}{{{n_f}^2}})\] , where R is the Rydberg’s constant=$1.097 \times {10^7}$
Energy of electron in nth orbit = $ - \dfrac{{13.6{z^2}}}{{{n^2}}}$ . For hydrogen z=1.
Complete step-by-step solution -
The energy difference between any two levels of the atom should be less than or equal to the energy of the electronic beam = 12.9 eV.
This can be formulated as $ - \dfrac{{13.6}}{{{n_f}^2}} + \dfrac{{13.6}}{{{n_i}^2}} \leqslant 12.9$ ni = 1.
$
- \dfrac{{13.6}}{{{n_f}^2}} + 13.6 \leqslant 12.9 \\
13.6 - 12.9 \leqslant \dfrac{{13.6}}{{{n_f}^2}} \\
{n_f}^2 \leqslant 19.43 \\
{n_f} \leqslant 4.4 \\
{n_f} = 4 \\
$
Therefore, the H-atom excites up to energy level 4.
For calculating the wavelength using the Rydberg’s formula: \[\dfrac{1}{\lambda } = R(\dfrac{1}{{{n_i}^2}} - \dfrac{1}{{{n_f}^2}})\]
For first wavelength of Balmer series- ni = 2 and nf = 3
$
\dfrac{1}{\lambda } = 1.097 \times {10^7}[\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}] \\
\lambda = 6563\mathop {\text{A}}\limits^{\text{o}} \\
$
For first wavelength of Paschen series- ni = 3 and nf = 4
$
\dfrac{1}{\lambda } = 1.097 \times {10^7}[\dfrac{1}{{{3^2}}} - \dfrac{1}{{{4^2}}}] \\
\lambda = 18761\mathop {\text{A}}\limits^{\text{o}} \\
$
The hydrogen atom is excited to the 4th energy level.
The first wavelength of the Balmer series is 6563 angstroms.
The first wavelength of Paschen series is 18761 angstroms.
Note-The energy level excitation could also be determined by finding the difference between individual pairs of energy levels and then comparing them with the energy of the electronic beam. It is important to remember the energy level formula and Rydberg's formula.
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