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Answer :
The correct answer is c) n = 3.
To determine the level to which the electron is promoted after absorbing a photon with a wavelength of 97.3 nm, we can use the Bohr model of the hydrogen atom, which relates the energy levels of an electron to its quantum number n. The energy of a photon is given by the equation
[tex]\[ E = \frac{hc}{\lambda} \][/tex]
where [tex]\( E \)[/tex] is the energy of the photon, [tex]\( h \)[/tex] is Planck's constant, [tex]\( c \)[/tex] is the speed of light, and [tex]\( \lambda \)[/tex] is the wavelength of the photon.
Given the wavelength [tex]\( \lambda = 97.3 \times 10^{-9} \)[/tex] meters (or 97.3 nm), we can calculate the energy of the photon:
[tex]\[ E = \frac{(6.626 \times 10^{-34} \text{ J s})(3.00 \times 10^8 \text{ m/s})}{97.3 \times 10^{-9} \text{ m}} \][/tex]
[tex]\[ E = \frac{(6.626 \times 10^{-34} \text{ J s})(3.00 \times 10^8 \text{ m/s})}{97.3 \times 10^{-9} \text{ m}} \][/tex]
[tex]\[ E \approx \frac{(6.626 \times 10^{-34} \text{ J s})(3.00 \times 10^8 \text{ m/s})}{97.3 \times 10^{-9} \text{ m}} \][/tex]
[tex]\[ E \approx 2.01 \times 10^{-18} \text{ J} \][/tex]
This energy is absorbed by the electron, promoting it to a higher energy level. The energy levels of the hydrogen atom are given by the equation:
[tex]\[ E_n = -\frac{13.6 \text{ eV}}{n^2} \][/tex]
where [tex]\( E_n \)[/tex] is the energy of the electron at level n, and 13.6 eV is the Rydberg energy for the hydrogen atom. To find the level to which the electron is promoted, we set the energy of the photon equal to the difference in energy between the initial level (n = 1) and the final level (n = ?):
[tex]\[ E = E_{\text{final}} - E_{\text{initial}} \][/tex]
[tex]\[ E = -\frac{13.6 \text{ eV}}{n_{\text{final}}^2} - \left(-\frac{13.6 \text{ eV}}{1^2}\right) \][/tex]
[tex]\[ E = 13.6 \text{ eV} \left(\frac{1}{1^2} - \frac{1}{n_{\text{final}}^2}\right) \][/tex]
Since we are looking for the quantum number [tex]\( n_{\text{final}} \)[/tex], we rearrange the equation:
[tex]\[ \frac{E}{13.6 \text{ eV}} = 1 - \frac{1}{n_{\text{final}}^2} \][/tex]
[tex]\[ \frac{1}{n_{\text{final}}^2} = 1 - \frac{E}{13.6 \text{ eV}} \][/tex]
[tex]\[ n_{\text{final}}^2 = \frac{1}{1 - \frac{E}{13.6 \text{ eV}}} \][/tex]
First, we convert the energy from joules to electron volts [tex](1 eV = 1.602 \times 10^{-19} J):[/tex]
[tex]\[ E \approx 2.01 \times 10^{-18} \text{ J} \times \frac{1 \text{ eV}}{1.602 \times 10^{-19} \text{ J}} \approx 12.55 \text{ eV} \][/tex]
Now we can find [tex]\( n_{\text{final}} \):[/tex]
[tex]\[ n_{\text{final}}^2 = \frac{1}{1 - \frac{12.55 \text{ eV}}{13.6 \text{ eV}}} \][/tex]
[tex]\[ n_{\text{final}}^2 = \frac{1}{1 - 0.923} \][/tex]
[tex]\[ n_{\text{final}}^2 = \frac{1}{0.077} \][/tex]
[tex]\[ n_{\text{final}}^2 \approx 12.987 \][/tex]
Taking the square root of both sides gives us:
[tex]\[ n_{\text{final}} \approx 3.6 \][/tex]
Since the quantum number n must be an integer, and the electron cannot be promoted to a fractional level, we round down to the nearest whole number, which is 3. Therefore, the electron is promoted to the n = 3 level."
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Final answer:
Promotion of electron in the hydrogen atom
When light with a wavelength of 97.3 nm is absorbed by an electron in the hydrogen atom, it promotes the electron to the n = 3 level.
Explanation:
When light with a wavelength of 97.3 nm is absorbed by an electron in the hydrogen atom, it promotes the electron to a higher energy level. The energy levels in the hydrogen atom are quantized and are given by the equation:
E = -13.6 eV / n^2
where n is the principal quantum number. To determine the level to which the electron is promoted, we need to find the value of n for which the absorbed light corresponds to the energy difference between the initial and final energy levels. Rearranging the equation, we get:
n = sqrt(-13.6 eV / E)
Substituting the wavelength of the absorbed light (97.3 nm) into the equation E = hc/λ, where h is Planck's constant (6.626 × 10^-34 J·s) and c is the speed of light (3.00 × 10^8 m/s), we can calculate the energy of the absorbed light. Plugging in the values, we find that the energy of the absorbed light is approximately 2.03 eV. Substituting this value into the equation for n, we find that n is approximately 3. Therefore, the electron is promoted to the n = 3 level.
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