JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Modern Physics

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 e V

. 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 ____.
Mass of the electron

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

and

1.0 e V = 1.6 \times 1 0^{- 19} J

Answer: 24

Solution

The number of photoelectrons emitted per second is

N = \frac{200}{6.25 \times 1.6 \times 1 0 ^{- 19}} = 2 \times 1 0^{20}

. Each electron gains kinetic energy

e V = 500 \times 1.6 \times 1 0^{- 19}

J, giving momentum

p = 2 m_{e} e V = 2 \times 9 \times 1 0^{- 31} \times 500 \times 1.6 \times 1 0^{- 19} = 1.2 \times 1 0^{- 23}

kg·m/s. The force on the anode is

F = N p = 2 \times 1 0^{20} \times 1.2 \times 1 0^{- 23} = 24 \times 1 0^{- 4}

N, so

n = 24

JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Numerical answer

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JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Modern Physics

Solution

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