Question

An electron and proton are separated by a large distance. The electron starts approaching the proton with energy 3 eV. The proton captures the electron and forms a hydrogen atom in second excited state. The resulting photon is incident on a photosensitive metal of threshold wavelength 4000 Å. What is the maximum kinetic energy of the emitted photoelectron?

• A. 3.3 eV
• B. No photoelectron would be emitted
• C. 7.61 eV
• D. 1.41 eV

Hint:

Remember the energy level conventions: "second excited state" means the principal quantum number is n=3. Also, the formula $E(\text{eV}) \approx 1240/\lambda(\text{nm})$ is a very useful shortcut for photoelectric effect and atomic physics problems.

Answer 1

The Correct Option is D

This problem involves three steps: finding the energy of the emitted photon, finding the work function of the metal, and finally using the photoelectric effect equation.
Step 1: Find the energy of the photon emitted during electron capture.
The initial energy of the system is the kinetic energy of the electron, $E_i = 3$ eV (potential energy is zero at large separation).
The electron is captured into the second excited state of the hydrogen atom. The ground state is n=1, first excited state is n=2, so the second excited state is n=3.
The energy of the n-th state is $E_n = -\frac{13.6}{n^2}$ eV.
The final energy of the hydrogen atom is $E_f = E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \approx -1.51$ eV.
By conservation of energy, $E_i = E_f + E_{photon}$.
$3 \text{ eV} = -1.51 \text{ eV} + E_{photon}$.
$E_{photon} = 3 + 1.51 = 4.51$ eV.
Step 2: Find the work function ($\phi$) of the photosensitive metal.
The work function is the minimum energy required to eject an electron, corresponding to the threshold wavelength $\lambda_0$.
$\lambda_0 = 4000$ Å = 400 nm.
Using the formula $E(\text{eV}) = \frac{1240 \text{ eV} \cdot \text{nm}}{\lambda(\text{nm})}$:
$\phi = \frac{1240}{400} = 3.1$ eV.
Step 3: Calculate the maximum kinetic energy ($K.E._{max}$) of the photoelectron.
According to the photoelectric effect equation: $K.E._{max} = E_{photon} - \phi$.
$K.E._{max} = 4.51 \text{ eV} - 3.1 \text{ eV} = 1.41$ eV.