To solve the problem step by step, we will follow the outlined process to determine the maximum kinetic energy of the emitted photoelectron.
### Step 1: Determine the Initial Energy of the System
The initial energy of the system consists of the energy of the approaching electron. Since the proton is at rest and the distance is large, we can ignore its potential energy at this stage.
\[
E_{\text{initial}} = E_{\text{electron}} = 3 \, \text{eV}
\]
### Step 2: Determine the Final Energy of the System
When the electron is captured by the proton, it forms a hydrogen atom. The energy of the electron in the second excited state (n=3) of hydrogen is given by the formula:
\[
E_n = -\frac{13.6 \, \text{eV}}{n^2}
\]
For the second excited state (n=3):
\[
E_3 = -\frac{13.6 \, \text{eV}}{3^2} = -\frac{13.6 \, \text{eV}}{9} = -1.51 \, \text{eV}
\]
Thus, the final energy of the system when the electron is captured is:
\[
E_{\text{final}} = E_3 + E_{\text{photon}} = -1.51 \, \text{eV} + E_{\text{photon}}
\]
### Step 3: Apply Conservation of Energy
According to the conservation of energy:
\[
E_{\text{initial}} = E_{\text{final}}
\]
Substituting the values we have:
\[
3 \, \text{eV} = -1.51 \, \text{eV} + E_{\text{photon}}
\]
Solving for \(E_{\text{photon}}\):
\[
E_{\text{photon}} = 3 \, \text{eV} + 1.51 \, \text{eV} = 4.51 \, \text{eV}
\]
### Step 4: Calculate the Work Function of the Metal
The threshold wavelength (\(\lambda_0\)) of the metal is given as \(4000 \, \text{Å}\). We can convert this to meters:
\[
\lambda_0 = 4000 \, \text{Å} = 4000 \times 10^{-10} \, \text{m} = 4 \times 10^{-7} \, \text{m}
\]
Using the equation for the energy of a photon:
\[
E = \frac{hc}{\lambda}
\]
Where \(h = 12400 \, \text{eV} \cdot \text{Å} / \text{m}\) and \(c = 3 \times 10^8 \, \text{m/s}\):
\[
\phi = \frac{12400 \, \text{eV} \cdot \text{Å}}{4000 \, \text{Å}} = 3.1 \, \text{eV}
\]
### Step 5: Apply Einstein's Photoelectric Equation
According to Einstein's photoelectric equation:
\[
K.E. = E_{\text{photon}} - \phi
\]
Substituting the values we found:
\[
K.E. = 4.51 \, \text{eV} - 3.1 \, \text{eV} = 1.41 \, \text{eV}
\]
### Conclusion
The maximum kinetic energy of the emitted photoelectron is:
\[
\boxed{1.41 \, \text{eV}}
\]
