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QPages 365-366: Q28: HC Verma: Concepts of Physics - Volume 2 Photoelectric Effect and Wave-Particle Duality Pages 365-366: Q28 Solution: In an experiment on photoelectric effect, the emitter and the…

In an experiment on photoelectric effect, the emitter and the collector plates are placed at a separation of 10 cm and are connected through an ammeter without any cell. A magnetic field BB exists parallel to the plates. The work function of the emitter is 2.39 eV and the light incident on it has wavelengths between 400 nm and 600 nm. Find the minimum value of BB for which the current registered by the ammeter is zero. Neglect any effect of space charge.

Step-by-Step Verified Solution

**Answer:** Bmin=2.8×10−5B_{min} = 2.8 \times 10^{-5} T. **Step-by-Step Explanation:** Maximum kinetic energy from 400 nm: KEmax=1240400−2.39=3.1−2.39=0.71KE_{max} = \frac{1240}{400} - 2.39 = 3.1 - 2.39 = 0.71 eV. For the current to be zero, the electron must not reach the collector. The magnetic force must provide centripetal force such that r=dr = d. v=2KEm=5.0×105v = \sqrt{\frac{2KE}{m}} = 5.0 \times 10^5 m/s. B=mved=9.1×10−31×5.0×1051.6×10−19×0.1=2.84×10−5B = \frac{mv}{ed} = \frac{9.1 \times 10^{-31} \times 5.0 \times 10^5}{1.6 \times 10^{-19} \times 0.1} = 2.84 \times 10^{-5} T.

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