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Photoelectric experiment

In a photoelectric experiment, a parallel beam of monochromati

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In a photoelectric experiment, a parallel beam of monochromatic light with power of $200 W$ is incident on a perfectly absorbing cathode of work function $6.25 eV$ . The frequency of light is just above the threshold frequency, so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is $100%$ . A potential difference of $500 V$ is applied between the cathode and the anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force $F = n \times 1 0^{- 4} N$ due to the impact of the electrons. The value of $n$ is ........... (Take mass of the electron, $m_{e} = 9 \times 1 0^{- 31} kg$ and $eV = 1.6 \times 1 0^{- 19} J$ )

Uploaded: 2022-07-16T13:56:48.000Z
Duration: 6 min

Views: 5,392 students

Updated on: Jul 16, 2022

Exam:

JEE Advanced 2018

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Emma Discussed

In a photoelectric experiment, a parallel beam of monochromatic light with power of

200 W

is incident on a perfectly absorbing cathode of work function

6.25 eV

. The frequency of light is just above the threshold frequency, so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is

100%

. A potential difference of

500 V

is applied between the cathode and the anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force

F = n \times 1 0^{- 4} N

due to the impact of the electrons. The value of

n

is ........... (Take mass of the electron,

m_{e} = 9 \times 1 0^{- 31} kg

and

eV = 1.6 \times 1 0^{- 19} J

)

12 mins ago

Discuss this question LIVE

12 mins ago

Text solutionVerified

∴ Power = nh f (where, n = number of photons incident per second) Since, KE = 0, h f = work-function W

KE = 0, h f = work-function W 200 = nW = n [6.25 \times 1.6 \times 1 0^{- 19}] n = \frac{200}{1.6 \times 1 0 ^{- 19} \times 6.25}

As photon is just above threshold frequency

K E_{m a x}

is zero and they are accelerated by potential difference of

500 V

KE_{f} \frac{P ^{2}}{2 m} = q Δ V = q Δ V \Rightarrow P = 2 m q Δ V

Since, efficiency is

100%

, number of electrons emitted per second

=

number of photons incident per second.

= n (mV) = n P = n 2 m q Δ V = \frac{200}{6.25 \times 1.6 \times 1 0 ^{- 19}} \times 2 (9 \times 1 0^{- 31}) \times 1.6 \times 1 0^{- 19} \times 500 = 24

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Question Text	In a photoelectric experiment, a parallel beam of monochromatic light with power of $200 W$ is incident on a perfectly absorbing cathode of work function $6.25 eV$ . The frequency of light is just above the threshold frequency, so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is $100%$ . A potential difference of $500 V$ is applied between the cathode and the anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force $F = n \times 1 0^{- 4} N$ due to the impact of the electrons. The value of $n$ is ........... (Take mass of the electron, $m_{e} = 9 \times 1 0^{- 31} kg$ and $eV = 1.6 \times 1 0^{- 19} J$ )
Updated On	Jul 16, 2022
Topic	Dual Nature of Radiation and Matter
Subject	Physics
Class	Class 12
Answer Type	Text solution:1 Video solution: 1
Upvotes	279
Avg. Video Duration	6 min

Text solutionVerified

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