Question

Ultraviolet light of wavelength 99 nm falls on a metal plate of work function 1.0 eV. If the mass of the electron is \( 9.1 \times 10^{-31} \) kg, the wavelength of the fastest photoelectron emitted is:

• A. 0.63 mm
• B. 0.66 nm
• C. 0.33 nm
• D. 0.36 nm

Hint:

Use the relationship \(E = \frac{hc}{\lambda}\) for photon energy and \(E_k = E - \phi\) for photoelectron energy.

Answer 1

The Correct Option is D

Step 1: Use the photoelectric equation

The photoelectric effect is described by the equation: \[ E_{\text{photon}} = W + E_{\text{kinetic}} \] where: - \( E_{\text{photon}} \) is the energy of the incident photon, - \( W \) is the work function of the metal, - \( E_{\text{kinetic}} \) is the kinetic energy of the emitted photoelectron. The energy of a photon is given by the equation: \[ E_{\text{photon}} = \frac{hc}{\lambda} \] where: - \( h \) is Planck’s constant (\( h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \)), - \( c \) is the speed of light (\( c = 3.0 \times 10^8 \, \text{m/s} \)), - \( \lambda \) is the wavelength of the light. The work function \( W \) is given as 1.0 eV, which we need to convert into joules: \[ W = 1.0 \, \text{eV} = 1.0 \times 1.602 \times 10^{-19} \, \text{J} = 1.602 \times 10^{-19} \, \text{J} \]

Step 2: Calculate the energy of the incident photon

The wavelength of the ultraviolet light is \( \lambda = 99 \, \text{nm} = 99 \times 10^{-9} \, \text{m} \). Using the formula for photon energy: \[ E_{\text{photon}} = \frac{(6.626 \times 10^{-34} \, \text{J} \cdot \text{s})(3.0 \times 10^8 \, \text{m/s})}{99 \times 10^{-9} \, \text{m}} \] Simplifying: \[ E_{\text{photon}} = \frac{1.9878 \times 10^{-25}}{99 \times 10^{-9}} = 2.01 \times 10^{-18} \, \text{J} \]

Step 3: Find the kinetic energy of the emitted photoelectron

From the photoelectric equation: \[ E_{\text{kinetic}} = E_{\text{photon}} - W \] Substituting the values: \[ E_{\text{kinetic}} = 2.01 \times 10^{-18} \, \text{J} - 1.602 \times 10^{-19} \, \text{J} = 1.85 \times 10^{-18} \, \text{J} \]

Step 4: Calculate the wavelength of the fastest photoelectron

The kinetic energy of the photoelectron is related to its momentum by: \[ E_{\text{kinetic}} = \frac{p^2}{2m} \] where \( p \) is the momentum of the photoelectron and \( m \) is the mass of the electron (\( m = 9.1 \times 10^{-31} \, \text{kg} \)). Solving for \( p \): \[ p = \sqrt{2mE_{\text{kinetic}}} \] Substituting the values: \[ p = \sqrt{2(9.1 \times 10^{-31} \, \text{kg})(1.85 \times 10^{-18} \, \text{J})} \] \[ p = \sqrt{3.37 \times 10^{-48}} = 5.81 \times 10^{-24} \, \text{kg} \cdot \text{m/s} \] The wavelength \( \lambda_{\text{electron}} \) of the fastest photoelectron is related to its momentum by the de Broglie relation: \[ \lambda_{\text{electron}} = \frac{h}{p} \] Substituting the values: \[ \lambda_{\text{electron}} = \frac{6.626 \times 10^{-34}}{5.81 \times 10^{-24}} = 1.14 \times 10^{-10} \, \text{m} = 0.114 \, \text{nm} \]

Final Answer:

The wavelength of the fastest photoelectron emitted is \( \boxed{0.114 \, \text{nm}} \).