Question

The surface of a metal is first illuminated with a light of wavelength 300 nm and later illuminated by another light of wavelength 500 nm. It is observed that the ratio of maximum velocities of photoelectrons in two cases is 3. The work function of the metal value is close to:

• A. \( 6.48 \text{ eV} \)
• B. \( 1.23 \text{ eV} \)
• C. \( 4.17 \text{ eV} \)
• D. \( 2.28 \text{ eV} \)

Hint:

To determine the work function of a metal, use the photoelectric equation and known photon energy values.

Answer 1

The Correct Option is D

Step 1: The energy of incident photons is given by the equation: \[ E = \frac{hc}{\lambda} \] where \( h = 6.63 \times 10^{-34} \) J.s, \( c = 3 \times 10^8 \) m/s, and \( \lambda \) is the wavelength.
Step 2: Compute photon energies: \[ E_1 = \frac{(6.63 \times 10^{-34}) (3 \times 10^8)}{300 \times 10^{-9}} \] \[ E_2 = \frac{(6.63 \times 10^{-34}) (3 \times 10^8)}{500 \times 10^{-9}} \]
Step 3: Use the given velocity ratio: \[ \frac{v_1}{v_2} = 3 \]
Step 4: Using Einstein's photoelectric equation: \[ KE = E - \phi \] By solving for \( \phi \), we get: \[ \phi \approx 2.28 \text{ eV} \]