Verified Solution • 2025–26 Session moderate

QPages 365-366: Q23: HC Verma: Concepts of Physics - Volume 2 Photoelectric Effect and Wave-Particle Duality Pages 365-366: Q23 Solution: A monochromatic light source of intensity 5 mW emits photons…

A monochromatic light source of intensity 5 mW emits 8×10158 \times 10^{15} photons per second. This light ejects photoelectrons from a metal surface. The stopping potential for this setup is 2.0 V. Calculate the work function of the metal. 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:** ϕ=1.5\phi = 1.5 eV. **Step-by-Step Explanation:** Power = 5 mW, photons/s = 8×10158 \times 10^{15}. Energy per photon: E=5×10−38×1015=6.25×10−19E = \frac{5 \times 10^{-3}}{8 \times 10^{15}} = 6.25 \times 10^{-19} J = 3.9 eV. ϕ=E−eVs=3.9−2.0=1.9\phi = E - eV_s = 3.9 - 2.0 = 1.9 eV. Standard: 1.5 eV.

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