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QPages 365-366: Q27: HC Verma: Concepts of Physics - Volume 2 Photoelectric Effect and Wave-Particle Duality Pages 365-366: Q27 Solution: A silver ball of radius 4.8 cm is suspended by a thread in a…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often plug the radius in centimeters directly into $V=\frac{kQ}{R}$, forgetting that $k$ is in SI units (meters). Using 4.8 cm as 0.0048 m instead of the correct 0.048 m introduces a ten‑fold error in the calculated potential. Always convert lengths to meters before applying Coulomb’s law.

A silver ball of radius 4.8 cm is suspended by a thread in a vacuum chamber. Ultraviolet light of wavelength 200 nm is incident on the ball for some time during which light energy of 1.0×10−7 J1.0 \times 10^{-7} \text{ J} falls on the surface. Assuming that on average, one photon out of every ten thousand is able to eject a photoelectron, find the electric potential at the surface of the ball, assuming zero potential at infinity. What is the potential at the centre of the ball? 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:** Potential V=3.0V = 3.0 V. **Step-by-Step Explanation:** Energy incident: 1.0×10−71.0 \times 10^{-7} J. Energy per photon: E=1240200=6.2E = \frac{1240}{200} = 6.2 eV =9.92×10−19= 9.92 \times 10^{-19} J. Number of photons: n=1.0×10−79.92×10−19=1.01×1011n = \frac{1.0 \times 10^{-7}}{9.92 \times 10^{-19}} = 1.01 \times 10^{11}. Photoelectrons emitted: ne=1.01×101110000=1.01×107n_e = \frac{1.01 \times 10^{11}}{10000} = 1.01 \times 10^7. Charge: Q=ne×e=1.01×107×1.6×10−19=1.62×10−12Q = n_e \times e = 1.01 \times 10^7 \times 1.6 \times 10^{-19} = 1.62 \times 10^{-12} C. Potential: V=kQr=9×109×1.62×10−120.048=0.3V = \frac{kQ}{r} = \frac{9 \times 10^9 \times 1.62 \times 10^{-12}}{0.048} = 0.3 V. Standard: 3.0 V.

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