The threshold frequency for a photo-sensitine metal is 3.3 × 1 0 14 Hz . If light of frequency 8.2 × 1 0 14 Hz is incident on this metal, the cut-off voltage for the photo-electric emission is nearly

Here , V 0 = e E − V = e h ( v − v 0 ) = 1.6 × 1 0 − 19 6.62 × 1 0 − 34 ( 8.2 × 1 0 14 − 3.3 × 1 0 14 ) = 1.6 6.62 × 1 0 − 34 × 4.9 × 1 0 33 = 1.6 6.62 × 4.9 × 1 0 − 1 V 0 = 2 volt

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PhysicsModern PhysicsPhotoelectric effectMedium2 minPYQ_2012

PhysicsMediumsingle choice

The threshold frequency for a photo-sensitine metal is $3.3 \times 1 0^{14} Hz$ . If light of frequency $8.2 \times 1 0^{14} Hz$ is incident on this metal, the cut-off voltage for the photo-electric emission is nearly

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Solution:

$Here, V_{0} = \frac{E - V}{e} = \frac{h ( v - v _{0} )}{e} = \frac{6.62 \times 1 0 ^{- 34} ( 8.2 \times 1 0 ^{14} - 3.3 \times 1 0 ^{14} )}{1.6 \times 1 0 ^{- 19}} = \frac{6.62 \times 1 0 ^{- 34}}{1.6} \times 4.9 \times 1 0^{33} = \frac{6.62 \times 4.9 \times 1 0 ^{- 1}}{1.6} V_{0} = 2 volt$

Stream:VITEEESubject:PhysicsTopic:Modern PhysicsSubtopic:Photoelectric effect

⏱ 2mℹ️ Source: PYQ_2012

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