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PhysicsModern PhysicsPhotoelectric effectMedium2 minPYQ_2016

PhysicsMediumstatement

Radiation of wavelength $λ$ is incident on a photocell. The fastest emitted photoelectron has a speed $v$ . If the wavelength is changed to $\frac{3 λ}{4}$ , the speed of the fastest emitted photoelectron will be

Options:

Answer:

Solution:

$\frac{1}{2} {mv}^{2} = \frac{hc}{λ} - ϕ$ ...(i)

$\frac{1}{2} mv'^{2} = \frac{4}{3} \frac{hc}{λ} - ϕ$ ...(ii)

From Eqn. (i)

$\frac{hc}{λ} = \frac{1}{2} {mv}^{2} + ϕ$

On putting this Equation (ii)

$\frac{1}{2} {mv}^{' 2} = \frac{4}{3} (\frac{1}{2} {mv}^{2} + ϕ) - ϕ$

$v'^{2} = \frac{4}{3} v^{2} + \frac{2 ϕ}{3 m}$

$\Rightarrow v' > \sqrt{\frac{4}{3}} v$

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:Photoelectric effect

⏱ 2mℹ️ Source: PYQ_2016

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