The threshold frequency for certain metal is 3.3 × 10 14 Hz . If the light of frequency 8.2 × 10 14 Hz is incident on the metal, the cut-off voltage of the photoelectric current will be

From relation e V s = h v - v 0 or V s = threshold or cut off voltage = h e v - v 0 = 6.6 × 10 - 34 1.6 × 10 - 19 ( 8.2 - 3.3 ) × 10 14 = 6 .6 × 4 .9 × 10 - 1 1 .6 = 2 V

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PhysicsModern PhysicsPhotoelectric effectEasy2 minPYQ_2020

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The threshold frequency for certain metal is $3.3 \times 10^{14}$ $Hz$ . If the light of frequency $8.2 \times 10^{14}$ $Hz$ is incident on the metal, the cut-off voltage of the photoelectric current will be

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Answer:

Solution:

From relation

$e V_{s} = h (v - v_{0})$

or $V_{s} =$ threshold or cut off voltage

$= \frac{h}{e} (v - v_{0})$

$= \frac{6.6 \times 10^{- 34}}{1.6 \times 10^{- 19}} (8.2 - 3.3) \times 10^{14}$

$= \frac{6 .6 \times 4 .9 \times 10^{- 1}}{1 .6} = 2 V$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Modern PhysicsSubtopic:Photoelectric effect

⏱ 2mℹ️ Source: PYQ_2020

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