To solve the problem step by step, we will follow the outlined process in the video transcript.
### Step 1: Calculate the energy of the incident photon
The energy of a photon can be calculated using the formula:
\[
E = \frac{hc}{\lambda}
\]
Where:
- \( h \) (Planck's constant) = \( 4.14 \times 10^{-15} \, \text{eV·s} \)
- \( c \) (speed of light) = \( 3 \times 10^8 \, \text{m/s} \)
- \( \lambda \) (wavelength) = \( 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m} \)
Substituting the values:
\[
E = \frac{(4.14 \times 10^{-15} \, \text{eV·s})(3 \times 10^8 \, \text{m/s})}{400 \times 10^{-9} \, \text{m}} = \frac{1.242 \times 10^{-6} \, \text{eV·m}}{400 \times 10^{-9} \, \text{m}} = 3.1 \, \text{eV}
\]
### Step 2: Calculate the maximum kinetic energy of the emitted photoelectron
The maximum kinetic energy (\( K_{\text{max}} \)) of the emitted photoelectron can be calculated using the formula:
\[
K_{\text{max}} = E - \phi
\]
Where:
- \( E \) = energy of the photon = \( 3.1 \, \text{eV} \)
- \( \phi \) (work function) = \( 2.3 \, \text{eV} \)
Substituting the values:
\[
K_{\text{max}} = 3.1 \, \text{eV} - 2.3 \, \text{eV} = 0.8 \, \text{eV}
\]
### Step 3: Calculate the energy of the He\(^+\) ion in the third excited state
The energy of the He\(^+\) ion in the nth excited state can be calculated using the formula:
\[
E_n = -\frac{13.6 \, \text{eV} \cdot Z^2}{n^2}
\]
Where:
- \( Z \) = atomic number of helium = 2
- \( n \) = principal quantum number for the third excited state = 4
Substituting the values:
\[
E_4 = -\frac{13.6 \, \text{eV} \cdot 2^2}{4^2} = -\frac{13.6 \, \text{eV} \cdot 4}{16} = -3.4 \, \text{eV}
\]
### Step 4: Calculate the energy of the emitted photon during the combination
The energy of the photon emitted during the combination can be calculated using the formula:
\[
\Delta E = K_{\text{max}} - E_4
\]
Substituting the values:
\[
\Delta E = 0.8 \, \text{eV} - (-3.4 \, \text{eV}) = 0.8 \, \text{eV} + 3.4 \, \text{eV} = 4.2 \, \text{eV}
\]
### Final Answer
The energy of the photon emitted during the combination is \( 4.2 \, \text{eV} \).
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