Question

A light wave described by E = 60[\(\sin(3 \times 10^{15})t + \sin(12 \times 10^{15})t\)] (in SI units) falls on a metal surface of work function 2.8 eV. The maximum kinetic energy of ejected photoelectron is (approximately)________ eV. (h=\(6.6 \times 10^{-34}\) J.s. and e=\(1.6 \times 10^{-19}\) C)

• A. 3.8
• B. 5.1
• C. 6.0
• D. 7.8

Hint:

The photoelectric effect is a quantum phenomenon where one photon interacts with one electron.
If the incident light contains multiple frequencies, it's like having multiple types of photons.
The maximum kinetic energy of an ejected electron will always be determined by the highest energy (highest frequency) photon, as it provides the largest energy packet to the electron.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
A light wave composed of two different frequencies illuminates a metal surface.
We need to find the maximum possible kinetic energy of the electrons ejected due to the photoelectric effect.
This maximum kinetic energy will be determined by the photon with the highest energy (highest frequency).
Step 2: Key Formula or Approach:
1. The electric field of a sinusoidal wave is described by \(E = E_0 \sin(\omega t)\), where \(\omega\) is the angular frequency.
2. The energy of a photon is related to its angular frequency by \(E_{photon} = hf = \frac{h\omega}{2\pi}\).
3. Einstein's photoelectric equation relates photon energy, work function (\(\phi\)), and maximum kinetic energy (\(K_{max}\)): \(K_{max} = E_{photon} - \phi\).
Step 3: Detailed Explanation:
The incident electric field is a superposition of two waves with angular frequencies:
\(\omega_1 = 3 \times 10^{15}\) rad/s.
\(\omega_2 = 12 \times 10^{15}\) rad/s.
To get the maximum kinetic energy of the photoelectrons, we must consider the incident photons with the highest energy. This corresponds to the highest frequency, so we use \(\omega_{max} = \omega_2 = 12 \times 10^{15}\) rad/s.
First, calculate the energy of the most energetic photons in Joules.
\[ E_{max_photon} = \frac{h\omega_{max}}{2\pi} = \frac{(6.6 \times 10^{-34} \text{ J.s}) \times (12 \times 10^{15} \text{ s}^{-1})}{2 \times 3.14159}
\] \[ E_{max_photon} = \frac{79.2 \times 10^{-19}}{6.283} \approx 12.605 \times 10^{-19} \text{ J}.
\] Next, convert this energy to electron-volts (eV) by dividing by the elementary charge \(e\).
\[ E_{max_photon} (\text{in eV}) = \frac{12.605 \times 10^{-19} \text{ J}}{1.6 \times 10^{-19} \text{ J/eV}} \approx 7.88 \text{ eV}.
\] The work function of the metal is given as \(\phi = 2.8\) eV.
Now, use Einstein's photoelectric equation to find the maximum kinetic energy.
\[ K_{max} = E_{max_photon} - \phi
\] \[ K_{max} = 7.88 \text{ eV} - 2.8 \text{ eV} = 5.08 \text{ eV}.
\] This value is approximately 5.1 eV.
Step 4: Final Answer:
The maximum kinetic energy of the ejected photoelectron is approximately 5.1 eV.