Question
An electron in an excited level of energy -21.4 eV falls to the ground level of energy -24.6 eV. Determine the wavelength of the photon emitted in the process.
[
1
e
V
=
1.6
×
10
-
19
;
6.6
×
10
-
34
J
s
]
Question 7Select one:
3.9
×
10
-
7
m
2.8
×
10
-
7
m
1.7
×
10
-
7
m
5.6
×
10
-
7
m
[
1
e
V
=
1.6
×
10
-
19
;
6.6
×
10
-
34
J
s
]
Question 7Select one:
3.9
×
10
-
7
m
2.8
×
10
-
7
m
1.7
×
10
-
7
m
5.6
×
10
-
7
m
Answers
GPT-4o mini
To find the wavelength of the photon emitted when the electron falls from an excited energy level to the ground state, we first need to calculate the energy difference between the two levels.
1. **Calculate the energy change (ΔE):**
\[
ΔE = E_{\text{final}} - E_{\text{initial}}
= (-24.6 \, \text{eV}) - (-21.4 \, \text{eV})
= -24.6 + 21.4
= -3.2 \, \text{eV}
\]
The negative sign indicates that energy is released (a photon is emitted).
2. **Convert energy from eV to Joules:**
\[
ΔE = 3.2 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV}
= 5.12 \times 10^{-19} \, \text{J}
\]
3. **Use the energy-wavelength relationship:**
The energy of a photon is related to its wavelength (λ) by the equation:
\[
E = \frac{hc}{λ}
\]
where:
- \(h = 6.63 \times 10^{-34} \, \text{J s}\) (Planck's constant),
- \(c = 3.00 \times 10^8 \, \text{m/s}\) (speed of light).
Rearranging the formula to solve for λ:
\[
λ = \frac{hc}{E}
\]
4. **Substituting the known values:**
Substituting values for \(h\), \(c\), and \(E\):
\[
λ = \frac{(6.63 \times 10^{-34} \, \text{J s}) (3.00 \times 10^8 \, \text{m/s})}{5.12 \times 10^{-19} \, \text{J}}
\]
\[
λ = \frac{1.989 \times 10^{-25} \, \text{J m}}{5.12 \times 10^{-19} \, \text{J}}
\]
\[
λ \approx 3.88 \times 10^{-7} \, \text{m}
\]
or approximately 3.9 × 10^-7 m.
5. **Conclusion:**
The wavelength of the photon emitted in the process is approximately:
\[
\boxed{3.9 \times 10^{-7} \, \text{m}}
\]
1. **Calculate the energy change (ΔE):**
\[
ΔE = E_{\text{final}} - E_{\text{initial}}
= (-24.6 \, \text{eV}) - (-21.4 \, \text{eV})
= -24.6 + 21.4
= -3.2 \, \text{eV}
\]
The negative sign indicates that energy is released (a photon is emitted).
2. **Convert energy from eV to Joules:**
\[
ΔE = 3.2 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV}
= 5.12 \times 10^{-19} \, \text{J}
\]
3. **Use the energy-wavelength relationship:**
The energy of a photon is related to its wavelength (λ) by the equation:
\[
E = \frac{hc}{λ}
\]
where:
- \(h = 6.63 \times 10^{-34} \, \text{J s}\) (Planck's constant),
- \(c = 3.00 \times 10^8 \, \text{m/s}\) (speed of light).
Rearranging the formula to solve for λ:
\[
λ = \frac{hc}{E}
\]
4. **Substituting the known values:**
Substituting values for \(h\), \(c\), and \(E\):
\[
λ = \frac{(6.63 \times 10^{-34} \, \text{J s}) (3.00 \times 10^8 \, \text{m/s})}{5.12 \times 10^{-19} \, \text{J}}
\]
\[
λ = \frac{1.989 \times 10^{-25} \, \text{J m}}{5.12 \times 10^{-19} \, \text{J}}
\]
\[
λ \approx 3.88 \times 10^{-7} \, \text{m}
\]
or approximately 3.9 × 10^-7 m.
5. **Conclusion:**
The wavelength of the photon emitted in the process is approximately:
\[
\boxed{3.9 \times 10^{-7} \, \text{m}}
\]