Verified Solution • 2025–26 Session moderate

QPages 365-366: Q1: HC Verma: Concepts of Physics - Volume 2 Photoelectric Effect and Wave-Particle Duality Pages 365-366: Q1 Solution: Visible light has wavelengths in the range of 400 nm to 780 nm. …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently assign the maximum photon energy to the longest wavelength (780 nm) due to linear intuition, but since $E = hc/\lambda$ is an inverse relationship, the maximum energy corresponds to the minimum wavelength (400 nm) and vice versa.

Visible light has wavelengths in the range of 400 nm to 780 nm. Calculate the range of energy of the photons of visible light. 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:** Range of energy: 1.59 eV to 3.10 eV. **Step-by-Step Explanation:** Energy of a photon: E=hcλ=1240 eV nmλE = \frac{hc}{\lambda} = \frac{1240 \text{ eV nm}}{\lambda}. For λ=780\lambda = 780 nm: E=1240/780=1.59E = 1240/780 = 1.59 eV. For λ=400\lambda = 400 nm: E=1240/400=3.10E = 1240/400 = 3.10 eV.

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