Question

If electric field of EM wave is given by $60[\sin(3 \times 10^{14} t) + \sin(12 \times 10^{14} t)]$ at $x = 0$, falls on a photosensitive material having work function $2.8$ eV. Find the maximum energy (in eV) of ejected electrons.

• A. $2.52$ eV
• B. $2.16$ eV
• C. $2.00$ eV
• D. $2.34$ eV

Hint:

In photoelectric effect problems with multiple frequencies:
Intensity affects number of electrons, not their energy.
Maximum kinetic energy depends only on the highest frequency present.

Answer 1

The Correct Option is B

Concept: From Einstein’s photoelectric equation, the maximum kinetic energy of the emitted electrons depends solely on the \textit{frequency} of the incident radiation and is independent of its intensity. \[ K_{\max} = h\nu - \phi \] When radiation contains more than one frequency component, the maximum kinetic energy is governed by the \textit{highest frequency} present in the incident electromagnetic wave.
Step 1: Determine frequencies from the given electric field The electric field is expressed as: \[ E = 60[\sin(3 \times 10^{14} t) + \sin(12 \times 10^{14} t)] \] Hence, the angular frequencies are: \[ \omega_1 = 3 \times 10^{14}, \quad \omega_2 = 12 \times 10^{14} \] The maximum angular frequency is: \[ \omega_{\max} = 12 \times 10^{14} \]
Step 2: Convert angular frequency to ordinary frequency \[ \nu = \frac{\omega}{2\pi} \] \[ \nu_{\max} = \frac{12 \times 10^{14}}{2\pi} \]
Step 3: Compute the photon energy Using Planck’s constant $h = 4.14 \times 10^{-15}$ eV·s: \[ E_{\max} = h\nu_{\max} = \frac{4.14 \times 10^{-15} \times 12 \times 10^{14}}{2\pi} \] \[ E_{\max} \approx 4.96 \text{ eV} \]
Step 4: Apply Einstein’s photoelectric equation Given the work function: \[ \phi = 2.8 \text{ eV} \] \[ K_{\max} = E_{\max} - \phi = 4.96 - 2.8 \] \[ K_{\max} = 2.16 \text{ eV} \] Conclusion: \[ \boxed{K_{\max} = 2.16 \text{ eV}} \]