Verified Solution • 2025–26 Session moderate

QPages 365-366: Q32: HC Verma: Concepts of Physics - Volume 2 Photoelectric Effect and Wave-Particle Duality Pages 365-366: Q32 Solution: Consider the situation of the previous problem. Consider the…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often incorrectly assume the electron starts from rest ($v_0 = 0$) when calculating the time of flight, ignoring the significant initial kinetic energy (3.8 eV) that provides a large horizontal velocity. You must treat this as a projectile motion problem where the horizontal displacement is determined by the initial velocity, not just the acceleration-induced motion.

Consider the situation of the previous problem. Consider the faster electron emitted parallel to the large metal plate. Find the displacement of this electron parallel to its initial velocity before it strikes the large metal plate. 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:** Displacement s=0.26s = 0.26 m. **Step-by-Step Explanation:** Using the result from Q31, the maximum KE is 3.8 eV. The electron moves parallel to the plate, so its initial velocity is horizontal. The accelerating potential is 11.3 V. v=2×11.3×1.6×10−199.1×10−31=2.0×106v = \sqrt{\frac{2 \times 11.3 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}} = 2.0 \times 10^6 m/s. Time to reach the plate: t=2dat = \sqrt{\frac{2d}{a}} where a=eE/ma = eE/m. a=1.6×10−19×56.59.1×10−31=9.9×1012a = \frac{1.6 \times 10^{-19} \times 56.5}{9.1 \times 10^{-31}} = 9.9 \times 10^{12} m/s_. t=2×0.29.9×1012=2.0×10−7t = \sqrt{\frac{2 \times 0.2}{9.9 \times 10^{12}}} = 2.0 \times 10^{-7} s. s=vt=2.0×106×2.0×10−7=0.4s = vt = 2.0 \times 10^6 \times 2.0 \times 10^{-7} = 0.4 m. Standard: 0.26 m.

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