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Dual Nature of Radiation question

2018 · Shift 2 · Q48
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Dual Nature of Radiation question

2018 · Shift 2 · Q48

JEE AdvancedPhysicsDual Nature of RadiationNumerical+3 / −1
In a photoelectric experiment a parallel beam of monochromatic light with power of 200W200W200W is incident on a perfectly absorbing cathode of work function 6.25ev.6.25ev.6.25ev. 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%100\%100%. A potential difference of 500V500V500V 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×10−4NF = n \times {10^{ - 4}}NF=n×10−4N due to the impact of the electrons. The value of nnn is ‾\underline{\hspace{2cm}}​. Mass of the electron Me=9×10−31 kg{M_e} = 9 \times {10^{ - 31}}\,kgMe​=9×10−31kg and 1.0eV=1.6×10−19 J.1.0eV = 1.6 \times {10^{ - 19}}\,J.1.0eV=1.6×10−19J.
Numerical answer
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Correct answer: 24

  1. Energy of each incident photon

Since the light frequency is just above threshold, the emitted photoelectrons have negligible initial kinetic energy. So photon energy is essentially equal to the work function:

Eγ=ϕ=6.25 eVE_\gamma = \phi = 6.25\,\text{eV}Eγ​=ϕ=6.25eV

Converting to joules:

Eγ=6.25×1.6×10−19=10×10−19=10−18 JE_\gamma = 6.25 \times 1.6 \times 10^{-19} = 10 \times 10^{-19} = 10^{-18}\,\text{J}Eγ​=6.25×1.6×10−19=10×10−19=10−18J

  1. Number of photons incident per second

Given power of the beam:

P=200 W=200 J s−1P = 200\,\text{W} = 200\,\text{J s}^{-1}P=200W=200J s−1

Number of photons per second:

N=PEγ=20010−18=2×1020 s−1N = \frac{P}{E_\gamma} = \frac{200}{10^{-18}} = 2 \times 10^{20}\,\text{s}^{-1}N=Eγ​P​=10−18200​=2×1020s−1

  1. Number of emitted electrons per second

Photoelectric efficiency is 100%100\%100%, so one photon emits one electron. Hence electrons emitted per second:

Ne=2×1020 s−1N_e = 2 \times 10^{20}\,\text{s}^{-1}Ne​=2×1020s−1

  1. Speed of each electron on reaching the anode

A potential difference of 500 V500\,\text{V}500V accelerates each electron. Since initial kinetic energy is negligible,

12mev2=eV\frac{1}{2}m_ev^2 = eV21​me​v2=eV

Now,

eV=500 eV=500×1.6×10−19=8×10−17 JeV = 500\,\text{eV} = 500 \times 1.6 \times 10^{-19} = 8 \times 10^{-17}\,\text{J}eV=500eV=500×1.6×10−19=8×10−17J

Thus,

12(9×10−31)v2=8×10−17\frac{1}{2}(9 \times 10^{-31})v^2 = 8 \times 10^{-17}21​(9×10−31)v2=8×10−17

v2=16×10−179×10−31=169×1014v^2 = \frac{16 \times 10^{-17}}{9 \times 10^{-31}} = \frac{16}{9} \times 10^{14}v2=9×10−3116×10−17​=916​×1014

v=43×107 m/sv = \frac{4}{3} \times 10^7\,\text{m/s}v=34​×107m/s

  1. Momentum delivered by each electron

Each electron is absorbed by the anode, so momentum transferred per electron is:

p=mev=9×10−31×43×107=12×10−24=1.2×10−23 kg m/sp = m_ev = 9 \times 10^{-31} \times \frac{4}{3} \times 10^7 = 12 \times 10^{-24} = 1.2 \times 10^{-23}\,\text{kg m/s}p=me​v=9×10−31×34​×107=12×10−24=1.2×10−23kg m/s

  1. Force on the anode

Force equals rate of change of momentum:

F=Ne⋅pF = N_e \cdot pF=Ne​⋅p

F=(2×1020)(1.2×10−23)F = (2 \times 10^{20})(1.2 \times 10^{-23})F=(2×1020)(1.2×10−23)

F=2.4×10−3 NF = 2.4 \times 10^{-3}\,\text{N}F=2.4×10−3N

Given

F=n×10−4 NF = n \times 10^{-4}\,\text{N}F=n×10−4N

So,

2.4×10−3=24×10−42.4 \times 10^{-3} = 24 \times 10^{-4}2.4×10−3=24×10−4

Hence,

n=24n = 24n=24

  1. Comparison with stored answer

Derived answer is 242424, which matches the stored correct answer.

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