To solve the problem step by step, we will follow the outlined process:
### Step 1: Calculate the Energy of the Incident Radiation
We start by calculating the energy of the incident radiation using the formula:
\[
E = \frac{hc}{\lambda}
\]
where:
- \( h = 4.14 \times 10^{-15} \, \text{eV-s} \) (Planck's constant)
- \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light)
- \( \lambda = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m} \)
Calculating \( E \):
\[
E = \frac{(4.14 \times 10^{-15} \, \text{eV-s}) \times (3 \times 10^8 \, \text{m/s})}{400 \times 10^{-9} \, \text{m}} = \frac{1.242 \times 10^{-6} \, \text{eV-m}}{400 \times 10^{-9} \, \text{m}} = 3.105 \, \text{eV}
\]
### Step 2: Calculate the Maximum Kinetic Energy of the Photoelectrons
The maximum kinetic energy \( K.E. \) of the emitted photoelectrons can be calculated using:
\[
K.E. = E - W
\]
where \( W \) is the work function given as \( 1.9 \, \text{eV} \).
Calculating \( K.E. \):
\[
K.E. = 3.105 \, \text{eV} - 1.9 \, \text{eV} = 1.205 \, \text{eV}
\]
### Step 3: Energy Released During the Combination
When the maximum energy electron combines with an alpha particle to form a \( \text{He}^+ \) ion, the energy released can be calculated as:
\[
E_{\text{released}} = K.E. + E_{He^+}
\]
The energy of the \( \text{He}^+ \) ion in the fourth excited state is given by:
\[
E_n = -\frac{13.6 \, Z^2}{n^2} \quad \text{(for hydrogen-like atoms)}
\]
For \( \text{He}^+ \) (where \( Z = 2 \) and \( n = 5 \)):
\[
E_5 = -\frac{13.6 \times 2^2}{5^2} = -\frac{54.4}{25} = -2.176 \, \text{eV}
\]
The energy released during the combination is:
\[
E_{\text{released}} = 1.205 \, \text{eV} - (-2.176 \, \text{eV}) = 1.205 + 2.176 = 3.381 \, \text{eV}
\]
### Step 4: Calculate Energy Levels for Transitions
Now we calculate the energy levels for transitions from the fourth excited state to lower states:
- \( E_4 = -\frac{13.6 \times 2^2}{4^2} = -\frac{54.4}{16} = -3.4 \, \text{eV} \)
- \( E_3 = -\frac{13.6 \times 2^2}{3^2} = -\frac{54.4}{9} = -6.04 \, \text{eV} \)
- \( E_2 = -\frac{13.6 \times 2^2}{2^2} = -\frac{54.4}{4} = -13.6 \, \text{eV} \)
### Step 5: Calculate the Energy of the Photons Emitted
The energy of the emitted photons during transitions can be calculated as:
1. From \( E_5 \) to \( E_4 \):
\[
E_{5 \to 4} = E_5 - E_4 = -2.176 - (-3.4) = 1.224 \, \text{eV}
\]
2. From \( E_5 \) to \( E_3 \):
\[
E_{5 \to 3} = E_5 - E_3 = -2.176 - (-6.04) = 3.864 \, \text{eV}
\]
3. From \( E_5 \) to \( E_2 \):
\[
E_{5 \to 2} = E_5 - E_2 = -2.176 - (-13.6) = 11.424 \, \text{eV}
\]
4. From \( E_4 \) to \( E_3 \):
\[
E_{4 \to 3} = E_4 - E_3 = -3.4 - (-6.04) = 2.64 \, \text{eV}
\]
5. From \( E_4 \) to \( E_2 \):
\[
E_{4 \to 2} = E_4 - E_2 = -3.4 - (-13.6) = 10.2 \, \text{eV}
\]
### Step 6: Identify Photons in the 2 to 4 eV Range
From the calculated energies, the photons that lie in the 2 to 4 eV range are:
- \( 2.64 \, \text{eV} \)
- \( 3.864 \, \text{eV} \)
### Final Answer
The energies of the photons likely to be emitted during and after the combination are:
- \( 2.64 \, \text{eV} \)
- \( 3.864 \, \text{eV} \)
