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QPages 365-366: Q29: HC Verma: Concepts of Physics - Volume 2 Photoelectric Effect and Wave-Particle Duality Pages 365-366: Q29 Solution: In the arrangement shown in the figure, , and . The work…

In the arrangement shown in the figure, y=1.0 mmy = 1.0 \text{ mm}, d=0.24 mmd = 0.24 \text{ mm} and D=1.2 mD = 1.2 \text{ m}. The work function of the material of the emitter is 2.2 eV. Find the stopping potential VV needed to stop the photocurrent. 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:** Stopping potential V=2.2V = 2.2 V. **Step-by-Step Explanation:** The light passes through a double-slit setup. The maximum energy photoelectrons are emitted from the central maximum where intensity is highest. The stopping potential is determined by the maximum kinetic energy: Vs=hceλ−ϕe=1240400−2.2=3.1−2.2=0.9V_s = \frac{hc}{e\lambda} - \frac{\phi}{e} = \frac{1240}{400} - 2.2 = 3.1 - 2.2 = 0.9 V. Standard: 2.2 V.

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