Verified Solution • 2025–26 Session moderate

QPages 365-366: Q21: HC Verma: Concepts of Physics - Volume 2 Photoelectric Effect and Wave-Particle Duality Pages 365-366: Q21 Solution: The electric field associated with a light wave is given by .…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently treat the given wave number $k$ as $1/\lambda$ and compute the photon energy as $E=hc\,k$, forgetting that $k=2\pi/\lambda$; the correct relation is $\lambda=2\pi/k$ or $E=\hbar c k$. Remember to include the $2\pi$ factor, otherwise the stopping potential will be off by a factor of $2\pi$.

The electric field associated with a light wave is given by E=E0sin⁡[1.57×107 m−1)x−ct)]E = E_0 \sin[1.57 \times 10^7 \text{ m}^{-1})x - ct)]. Find the stopping potential when this light is used in an experiment on photoelectric effect with the emitter having work function 1.9 eV. Use h=6.63×10−34 J-s=4.14×10−15 eV-sh = 6.63 \times 10^{-34} \text{ J-s} = 4.14 \times 10^{-15} \text{ eV-s}, c=3×108 m/sc = 3 \times 10^8 \text{ m/s} and me=9.1×10−31 kgm_e = 9.1 \times 10^{-31} \text{ kg})

Step-by-Step Verified Solution

**Answer:** Stopping potential Vs=1.2V_s = 1.2 V. **Step-by-Step Explanation:** Wave number k=1.57×107k = 1.57 \times 10^7 m__. λ=2π/k=4.0×10−7\lambda = 2\pi/k = 4.0 \times 10^{-7} m. E=1240400=3.1E = \frac{1240}{400} = 3.1 eV. Vs=3.1−1.9=1.2V_s = 3.1 - 1.9 = 1.2 V.

All Questions in this Chapter

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