1. Energy of the incident electron before capture

The electron is initially far away from the proton, so we take the potential energy at infinity as zero.

Given kinetic energy of electron: $K = 3\ \text{eV}$

So total initial energy of the electron-proton system is: $E_i = 3\ \text{eV}$

2. Energy of hydrogen atom in second excited state

Second excited state means:

• ground state: $n=1$
• first excited state: $n=2$
• second excited state: $n=3$

Energy of hydrogen atom in level $n$ is: $E_n = -\frac{13.6}{n^2}\ \text{eV}$

For $n=3$: $E_3 = -\frac{13.6}{9} = -1.51\ \text{eV}$

3. Energy of emitted photon during capture

By conservation of energy, the emitted photon carries away the excess energy: $E_\gamma = E_i - E_f$ where $E_f = -1.51\ \text{eV}$.

Thus, $E_\gamma = 3 - (-1.51) = 4.51\ \text{eV}$

4. Work function of the metal

Threshold wavelength: $\lambda_0 = 4000\ \text{\AA} = 400\ \text{nm}$

Work function is: $\phi = \frac{hc}{\lambda_0}$

Using $hc = 1240\ \text{eV·nm}$: $\phi = \frac{1240}{400} = 3.1\ \text{eV}$

5. Maximum kinetic energy of photoelectron

Using Einstein’s photoelectric equation: $K_{\max} = E_\gamma - \phi$

So, $K_{\max} = 4.51 - 3.1 = 1.41\ \text{eV}$

6. Option check

• A: $7.61\ \text{eV}$ ❌
• B: $1.41\ \text{eV}$ ✅
• C: $3.3\ \text{eV}$ ❌
• D: No photoelectron would be emitted ❌ since $4.51 > 3.1$

Therefore, the correct option is: $\boxed{\text{B: }1.41\ \text{eV}}$

7. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

So they agree.