Question

If the electric field of an EM wave is given by \[ 60\,[\sin(3\times10^{14}t) + \sin(12\times10^{14}t)] \] at \( x = 0 \) and it falls on a photosensitive material having work function \( 2.8\,\text{eV} \), find the maximum kinetic energy (in eV) of the ejected electrons.

• A. \(2.52\,\text{eV}\)
• B. \(2.16\,\text{eV}\)
• C. \(2.00\,\text{eV}\)
• D. \(2.34\,\text{eV}\)

Hint:

For photoelectric effect problems:
Maximum kinetic energy depends only on the highest frequency
Intensity affects number of electrons, not their energy
Use \( K_{\max} = h\nu_{\max} - \phi \)

Answer 1

The Correct Option is B

Step 1: The given electric field consists of two frequencies: \[ \nu_1 = 3\times10^{14}\,\text{Hz}, \nu_2 = 12\times10^{14}\,\text{Hz} \] 
Step 2: In the photoelectric effect, only the highest frequency photon determines the maximum kinetic energy of emitted electrons. \[ \nu_{\max} = 12\times10^{14}\,\text{Hz} \] 
Step 3: Energy of a photon: \[ E = h\nu \] Using \( h = 4.14\times10^{-15}\,\text{eVs} \): \[ E_{\max} = 4.14\times10^{-15} \times 12\times10^{14} = 4.97\,\text{eV} \] 
Step 4: Apply Einstein’s photoelectric equation: \[ K_{\max} = E_{\max} - \phi = 4.97 - 2.8 = 2.17\,\text{eV} \] \[ \boxed{K_{\max} \approx 2.16\,\text{eV}} \]