Question

A light whose frequency is equal to \(6 \times 10^{14}\) Hz is incident on a metal whose work function is 2 eV. The maximum energy of the electrons emitted will be

• A. 2.49 eV
• B. 4.49 eV
• C. 0.49 eV
• D. 5.49 eV

Hint:

Maximum kinetic energy in photoelectric effect is \(K_{\max} = h\nu - \phi\).

Answer 1

The Correct Option is C

Step 1: Use Einstein photoelectric equation.
\[ K_{\max} = h\nu - \phi \]
Step 2: Substitute values.
Given:
\[ h = 6.63 \times 10^{-34} \, \text{J s}, \quad \nu = 6 \times 10^{14}\, \text{Hz} \]
Energy of photon:
\[ h\nu = 6.63 \times 10^{-34} \times 6 \times 10^{14} = 3.978 \times 10^{-19}\, \text{J} \]
Step 3: Convert joule to eV.
\[ 1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J} \]
\[ h\nu = \frac{3.978 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.49 \,\text{eV} \]
Step 4: Subtract work function.
\[ K_{\max} = 2.49 - 2 = 0.49 \,\text{eV} \]
Final Answer:
\[ \boxed{0.49 \,\text{eV}} \]