Question

Light of wavelength \( 4000\text{\AA} \) is incident on a sodium surface for which the threshold wavelength of photoelectrons is \( 5420\text{\AA} \). The work function of sodium is:

• A. \( 4.58 \text{ eV} \)
• B. \( 2.29 \text{ eV} \)
• C. \( 1.14 \text{ eV} \)
• D. \( 0.57 \text{ eV} \)

Hint:

The work function of a material is the minimum energy required to eject an electron and is calculated using the threshold wavelength.

Answer 1

The Correct Option is B

Step 1: Understanding the Photoelectric Equation
The energy of incident photons is given by the equation: \[ E = \frac{hc}{\lambda} \] where: - \( h = 6.626 \times 10^{-34} \text{ Js} \) (Planck's constant),
- \( c = 3.0 \times 10^8 \text{ m/s} \) (speed of light),
- \( \lambda \) is the wavelength of the incident light.
The work function \( \phi \) (minimum energy required to eject electrons) is given by: \[ \phi = \frac{hc}{\lambda_0} \] where \( \lambda_0 = 5420\text{\AA} = 5420 \times 10^{-10} \text{ m} \) is the threshold wavelength.
Step 2: Calculating the Work Function
Substituting the known values: \[ \phi = \frac{(6.626 \times 10^{-34}) (3.0 \times 10^8)}{5420 \times 10^{-10}} \] \[ = \frac{1.9878 \times 10^{-25}}{5.42 \times 10^{-7}} \] \[ = 3.67 \times 10^{-19} \text{ J} \] Converting to electron volts: \[ \phi = \frac{3.67 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.29 \text{ eV}. \] Final Answer: The work function of sodium is \( 2.29 \text{ eV} \), which matches option (2).