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.
Show 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.
Concept:
According to Einstein’s photoelectric equation, the maximum kinetic energy of emitted electrons depends only on the \textit{frequency} of the incident radiation and not on its intensity.
\[
K_{\max} = h\nu - \phi
\]
If multiple frequencies are present, the maximum kinetic energy is determined by the \textit{highest frequency} component of the incident electromagnetic wave.
Step 1: Identify frequencies from the given electric field
The electric field is:
\[
E = 60[\sin(3 \times 10^{14} t) + \sin(12 \times 10^{14} t)]
\]
Thus, the angular frequencies are:
\[
\omega_1 = 3 \times 10^{14}, \quad \omega_2 = 12 \times 10^{14}
\]
The higher angular frequency is:
\[
\omega_{\max} = 12 \times 10^{14}
\]
Step 2: Convert angular frequency to linear frequency
\[
\nu = \frac{\omega}{2\pi}
\]
\[
\nu_{\max} = \frac{12 \times 10^{14}}{2\pi}
\]
Step 3: Calculate photon energy
Using $h = 4.14 \times 10^{-15}$ 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 photoelectric equation
Given work function:
\[
\phi = 2.8 \text{ eV}
\]
\[
K_{\max} = E_{\max} - \phi = 4.96 - 2.8
\]
\[
K_{\max} = 2.16 \text{ eV}
\]
Conclusion:
\[
\boxed{K_{\max} = 2.16 \text{ eV}}
\]
