Light of frequency 5 v 0 falls on a photosensitive surface of the threshold frequency v 0 . The velocity of emitted electrons is 4 × 1 0 5 m / s . If now light of frequency 9 v 0 is incident on the surface, then velocity of electron will become:

We have 2 1 mv 2 = h v − h v 0 2 1 mv 1 2 = h ( 5 v 0 − v 0 ) ... ( 1 ) 2 1 mv 2 2 = h ( 9 v 0 − v 0 ) ... ( 2 ) ( 2 ) / ( 1 ) v 1 2 v 2 2 = 4 v 0 8 v 0 = 2 v 2 = 2 v 1 = 2 × 4 × 1 0 5 = 5.64 × 1 0 5 m / sec

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PhysicsModern PhysicsPhotoelectric effectMedium2 minPYQ_2023

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Light of frequency $5 v_{0}$ falls on a photosensitive surface of the threshold frequency $v_{0}$ . The velocity of emitted electrons is $4 \times 1 0^{5} m / s$ . If now light of frequency $9 v_{0}$ is incident on the surface, then velocity of electron will become:

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Answer:

Solution:

We have $\frac{1}{2} mv^{2} = h v - h v_{0}$ $\frac{1}{2} mv_{1}^{2} = h (5 v_{0} - v_{0}) ... (1) \frac{1}{2} mv_{2}^{2} = h (9 v_{0} - v_{0}) ... (2)$ $(2) / (1)$ $\frac{v _{2}^{2}}{v _{1}^{2}} = \frac{8 v _{0}}{4 v _{0}} = 2 v_{2} = 2 v_{1} = 2 \times 4 \times 1 0^{5} = 5.64 \times 1 0^{5} m / sec$

Stream:VITEEESubject:PhysicsTopic:Modern PhysicsSubtopic:Photoelectric effect

⏱ 2mℹ️ Source: PYQ_2023

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