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PhysicsModern PhysicsPhotoelectric effectEasy2 minPYQ_2020

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When light of wavelength $300 nm$ or less falls on a photoelectric emitter $A$ , photoelectrons are emitted. For another emitter $B$ , light of wavelength $600 nm$ is sufficient for releasing photoelectrons. The ratio of the work function of emitter $A$ to $B$ is

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Answer:

Solution:

Work function is given by

$ϕ = \frac{h c}{λ} o r ϕ \propto \frac{1}{λ}$

$∵ \frac{ϕ_{1}}{ϕ_{2}} = \frac{λ_{2}}{λ_{1}}$

$= \frac{600}{300} = \frac{2}{1}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Modern PhysicsSubtopic:Photoelectric effect

⏱ 2mℹ️ Source: PYQ_2020

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