Question

A metal has work function \(2.5\,\text{eV}\). If a radiation of frequency \(3.2\times10^{15}\,\text{Hz}\) is incident on this metal surface, then the maximum kinetic energy of ejected photoelectrons is \((h=6.6\times10^{-34}\,\text{J-s})\)

• A. \(9.5\,\text{eV}\)
• B. \(2.5\,\text{eV}\)
• C. \(10.7\,\text{eV}\)
• D. \(12.6\,\text{eV}\)

Hint:

In photoelectric effect, \[ K_{\max}=h\nu-\phi. \] Convert photon energy from joule to eV before subtracting the work function if the work function is given in eV.

Answer 1

The Correct Option is C

Step 1: Apply Einstein's photoelectric equation.
K_max = hν – φ.

Step 2: Compute photon energy.
hν = 6.6×10⁻³⁴ × 3.2×10¹⁵ = 21.12×10⁻¹⁹ J = 13.2 eV.

Step 3: Subtract work function.
K_max = 13.2 – 2.5 = 10.7 eV.

Step 4: Final Answer:
10.7 eV.