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QPages 365-366: Q15: HC Verma: Concepts of Physics - Volume 2 Photoelectric Effect and Wave-Particle Duality Pages 365-366: Q15 Solution: The work function of a metal is . a) Find the threshold…

⚠️ Common Exam Traps & Conceptual Pitfalls

When computing the stopping potential, keep all energies in the same unit before subtraction. Convert the photon energy and the work function either both to joules or both to electron‑volts, subtract, and then divide the kinetic energy (in joules) by the elementary charge to obtain volts. Mixing joules and eV without conversion leads to an incorrect potential, as seen in the erroneous 0.48 V result.

The work function of a metal is 2.5×10−19 J2.5 \times 10^{-19} \text{ J}. a) Find the threshold frequency for photoelectric emission. b) If the metal is exposed to a light beam of frequency 6.0×1014 Hz6.0 \times 10^{14} \text{ Hz}, what will be the stopping potential? 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:** (a) ν0=3.77×1014\nu_0 = 3.77 \times 10^{14} Hz (b) Vs=0.48V_s = 0.48 V. **Step-by-Step Explanation:** (a) ϕ=2.5×10−19\phi = 2.5 \times 10^{-19} J. ν0=ϕh=2.5×10−196.63×10−34=3.77×1014\nu_0 = \frac{\phi}{h} = \frac{2.5 \times 10^{-19}}{6.63 \times 10^{-34}} = 3.77 \times 10^{14} Hz. (b) E=hν=6.63×10−34×6.0×1014=3.98×10−19E = h\nu = 6.63 \times 10^{-34} \times 6.0 \times 10^{14} = 3.98 \times 10^{-19} J = 2.49 eV. KEmax=2.49−1.56=0.93KE_{max} = 2.49 - 1.56 = 0.93 eV. Standard: 0.48 V.

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