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PhysicsModern PhysicsPhotoelectric effectEasy2 minPYQ_2020

PhysicsEasysingle choice

$K_{1}$ be the maximum kinetic energy of photoelectrons emitted when metal surface is illuminated by light of wavelength $λ_{1}$ . Similarly $K_{2}$ is corresponding to wavelength $λ_{2}$ for same metal. If $λ_{1} = 2 λ_{2}$ then

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Answer:

Solution:

$K_{1} = \frac{hc}{λ_{1}} - W$ ....(i)

and $K_{2} = \frac{hc}{λ_{2}} - W$ ....(ii)

Substituting $λ_{1} = 2 λ_{2}$ in Eq. (i), we get

$K_{1} = \frac{hc}{2 λ_{2}} - W$

$= \frac{1}{2} (\frac{hc}{λ_{2}}) - W = \frac{1}{2} (K_{2} + W) - W$

$∴ K_{1} = \frac{K_{2}}{2} - \frac{W}{2}$ or $K_{1} < \frac{K_{2}}{2}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Modern PhysicsSubtopic:Photoelectric effect

⏱ 2mℹ️ Source: PYQ_2020

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