In a photoelectric set up the most energetic photoelectron from the material is introduced horizontally parallel to the parallel plate capacitor of length ℓ and constant electric field E, it gets defl

Question

In a photoelectric set up the most energetic photoelectron from the material is introduced horizontally parallel to the parallel plate capacitor of length $ℓ$ and constant electric field E, it gets deflected by d when emerges out from the capacitor. The stopping potential for this electron will be (neglecting the effect of gravity)

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Answer:

A

Solution:

$t = \frac{ℓ}{v} d = \frac{1}{2} \frac{e E}{m} \frac{ℓ ^{2}}{v ^{2}} \frac{1}{2} m v^{2} = \frac{e E ℓ ^{2}}{4 d} V_{s} = \frac{E ℓ ^{2}}{4 d}$

TestHub · Accepted Answer

​ t = v ℓ ​ ​ d = 2 1 ​ m e E ​ v 2 ℓ 2 ​ ​ 2 1 ​ m v 2 = 4 d e E ℓ 2 ​ ​ V s ​ = 4 d E ℓ 2 ​ ​

Physics - Modern Physics Question with Solution | TestHub

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