Question

In a photoelectric effect, the kinetic energy of electrons is \(1.3 \times 10^{-19}\,\text{J}\). If work function of the metal is \(2.27\,\text{eV}\), then the frequency (in Hz) of the incident radiation is (\(h = 6.6 \times 10^{-34}\,\text{J s}\), \(1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J}\)).

• A. \(7.47 \times 10^{14}\)
• B. \(7.47 \times 10^{15}\)
• C. \(6.47 \times 10^{15}\)
• D. \(3.47 \times 10^{14}\) \bigskip

Hint:

For photoelectric effect: \[ hf=\phi+K_{\max} \] Always convert the work function from eV to joules before substituting into the equation.

Answer 1

The Correct Option is B

Concept: Photoelectric equation: \(E = \phi + K_{\max}\). Find photon energy, then frequency.

Step 1: \(\phi = 2.27 \times 1.6\times10^{-19} = 3.632\times10^{-19}\) J. \(E = 3.632\times10^{-19} + 1.3\times10^{-19} = 4.932\times10^{-19}\) J. \(f = E/h = 4.932\times10^{-19} / 6.6\times10^{-34} = 7.47\times10^{14}\) Hz.

Step 2: Write the final answer. \(\boxed{f=7.47\times10^{14}\,\text{Hz}}\)