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

To solve this problem, we need to determine the maximum kinetic energy of the electrons ejected from the photosensitive material when an electromagnetic (EM) wave is incident upon it. The given electric field expression of the EM wave is:

\[ E(t) = 60 [\sin(3 \times 10^{14} t) + \sin(12 \times 10^{14} t)] \]

This electric field is composed of two sinusoidal components with different frequencies:

• \(\nu_1 = 3 \times 10^{14}\,\text{Hz}\)
• \(\nu_2 = 12 \times 10^{14}\,\text{Hz}\)

In photoelectric effect analysis, the energy of the incident photon is given by Planck's relation:

\[ E_{\text{photon}} = h \nu \]

where \(h\) is Planck's constant (\(6.626 \times 10^{-34}\,\text{Js}\)).

For the two frequency components, the photon energies are:

• \(E_{\text{photon, 1}} = h \times 3 \times 10^{14}\)
• \(E_{\text{photon, 2}} = h \times 12 \times 10^{14}\)

We will calculate the photon energies in electron-volts (eV):

• \[ E_{\text{photon, 1}} = 4.1357 \times 10^{-15}\,\text{eV s} \times 3 \times 10^{14}\,\text{Hz} = 1.24\,\text{eV} \]
• \[ E_{\text{photon, 2}} = 4.1357 \times 10^{-15}\,\text{eV s} \times 12 \times 10^{14}\,\text{Hz} = 4.96\,\text{eV} \]

The maximum kinetic energy (\(K_{\text{max}}\)) of ejected electrons is given by the photoelectric equation:

\[ K_{\text{max}} = E_{\text{photon}} - \phi \]

where \(\phi = 2.8\,\text{eV}\) is the work function of the material.

Since the first photon energy \(E_{\text{photon, 1}} = 1.24\,\text{eV}\) is less than the work function, it will not eject any electrons. Therefore, we consider the second photon energy, \(E_{\text{photon, 2}}\), for calculation:

\[ K_{\text{max}} = 4.96\,\text{eV} - 2.8\,\text{eV} = 2.16\,\text{eV} \]

Thus, the maximum kinetic energy of the ejected electrons is 2.16 eV, which corresponds to the given correct answer.