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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Photoelectric Effect and Wave–Particle Duality

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Photoelectric Effect and Wave–Particle Duality

In a photoelectric effect experiment, the stopping potential is...

In a photoelectric effect experiment, the stopping potential is measured for monochromatic light beams of varying wavelengths. Determine the work function of the emitter from the experimental data.
wavelength (nm)350350400400450450500500550550
stopping potential (V)1.451.451.001.000.660.660.380.380.160.16

Options

  1. A2.15 eV2.15 \text{ eV}
  2. B1.08 eV1.08 \text{ eV}
  3. C3.15 eV3.15 \text{ eV}
  4. D4.30 eV4.30 \text{ eV}

Correct answer

A. 2.15 eV2.15 \text{ eV}

Step-by-step solution

According to Einstein's photoelectric equation:

eVs=hcλ−ϕeV_s = \dfrac{hc}{\lambda} - \phi

where VsV_s is the stopping potential, λ\lambda is the wavelength of incident light, and ϕ\phi is the work function of the material.

Using the standard value hc≈1240 eV nmhc \approx 1240 \text{ eV nm}, we can calculate the work function for the given data points.

For λ=400 nm\lambda = 400 \text{ nm} and Vs=1.00 VV_s = 1.00 \text{ V}:

ϕ=1240400−1.00\phi = \dfrac{1240}{400} - 1.00

ϕ=3.10−1.00=2.10 eV\phi = 3.10 - 1.00 = 2.10 \text{ eV}

For λ=500 nm\lambda = 500 \text{ nm} and Vs=0.38 VV_s = 0.38 \text{ V}:

ϕ=1240500−0.38\phi = \dfrac{1240}{500} - 0.38

ϕ=2.48−0.38=2.10 eV\phi = 2.48 - 0.38 = 2.10 \text{ eV}

The calculated work function is approximately 2.10 eV2.10 \text{ eV}. Among the given choices, the closest value is 2.15 eV2.15 \text{ eV}.

Answer: 2.15 eV2.15 \text{ eV}
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