Question

Work function for an object is 2.3 eV. If maximum kinetic energy of ejected electrons is 0.18 eV, find wavelength \( \lambda \) of incident photon on object.

• A. 200 nm
• B. 500 nm
• C. 250 nm
• D. 300 nm

Hint:

Remember, the photoelectric equation relates the energy of the photon to the work function and the kinetic energy of ejected electrons. Use \( E_{\text{photon}} = \frac{h c}{\lambda} \) to find the wavelength.

Answer 1

The Correct Option is B

Step 1: Understanding the photoelectric equation.
The energy of an incident photon is related to its wavelength by the equation: \[ E_{\text{photon}} = \frac{h c}{\lambda} \] Where:
- \( h \) is Planck's constant \( = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \),
- \( c \) is the speed of light \( = 3 \times 10^8 \, \text{m/s} \),
- \( \lambda \) is the wavelength of the incident radiation.
According to Einstein’s photoelectric equation: \[ E_{\text{photon}} = \text{Work Function} + E_{\text{kinetic}} \] Where:
- Work function \( \phi = 2.3 \, \text{eV} \),
- Maximum kinetic energy of the emitted electron \( E_{\text{kinetic}} = 0.18 \, \text{eV} \).

Step 2: Calculating the total energy of the photon.
The total energy of the incident photon is obtained by adding the work function and the maximum kinetic energy: \[ E_{\text{photon}} = 2.3 \, \text{eV} + 0.18 \, \text{eV} \] \[ E_{\text{photon}} = 2.48 \, \text{eV} \]
Step 3: Converting energy from electron volts to joules.
Since \( 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \), the photon energy in joules is: \[ E_{\text{photon}} = 2.48 \times 1.602 \times 10^{-19} \] \[ E_{\text{photon}} = 3.97 \times 10^{-19} \, \text{J} \]
Step 4: Determining the wavelength of the photon.
Using the relation between energy and wavelength: \[ \lambda = \frac{h c}{E_{\text{photon}}} \] Substituting the known values: \[ \lambda = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{3.97 \times 10^{-19}} \] \[ \lambda = 5.0 \times 10^{-7} \, \text{m} \] \[ \lambda = 500 \, \text{nm} \]
Step 5: Conclusion.
Hence, the wavelength of the incident photon required to produce the given kinetic energy is \( 500 \, \text{nm} \).
Final Answer: \( 500 \, \text{nm} \).