JEE Challenger
More from Dual Nature of Radiation and Matter
PhyHardAdv 2018 • Paper 2

Force Experienced by Anode due to Photoelectron Impact

In a photoelectric setup, a parallel beam of monochromatic light of power 200 W200\text{ W} falls on a perfectly absorbing cathode having work function 6.25 eV6.25\text{ eV}. The frequency of light is slightly above the threshold value so photoelectrons leave with negligible kinetic energy. A potential difference of 500 V500\text{ V} is applied between cathode and anode. All emitted photoelectrons reach the anode and get absorbed. If the anode experiences a force F=n×10−4 NF = n \times 10^{-4}\text{ N} from electron impact, find nn. (Given me=9×10−31 kgm_e = 9 \times 10^{-31}\text{ kg} and 1.0 eV=1.6×10−19 J1.0\text{ eV} = 1.6 \times 10^{-19}\text{ J})

Official Numerical Answer24

Topics & Concepts

Dual Nature of Radiation and MatterPhotoelectric Effect

Step-by-Step Solution

To find the value of nn, we calculate the force exerted on the anode by the photoelectrons hitting and getting absorbed by it.

Step 1: Energy of the incident photons and rate of emission of photoelectrons

The energy of each photon is given to be just above the threshold energy (work function) of the cathode metal: Ephoton≈Φ=6.25 eVE_{\text{photon}} \approx \Phi = 6.25 \text{ eV}

Converting this energy to Joules: Ephoton=6.25×1.6×10−19 J=10−18 JE_{\text{photon}} = 6.25 \times 1.6 \times 10^{-19} \text{ J} = 10^{-18} \text{ J}

Given the total incident power P=200 WP = 200 \text{ W}, the rate of incident photons dNdt\frac{dN}{dt} is: dNdt=PEphoton=200 W10−18 J=2×1020 s−1\frac{dN}{dt} = \frac{P}{E_{\text{photon}}} = \frac{200 \text{ W}}{10^{-18} \text{ J}} = 2 \times 10^{20} \text{ s}^{-1}

Since the photoelectron emission efficiency is 100%100\%, the rate at which photoelectrons are emitted from the cathode (and thus reach the anode) is: dNedt=2×1020 s−1\frac{dN_e}{dt} = 2 \times 10^{20} \text{ s}^{-1}

Step 2: Momentum of photoelectrons upon reaching the anode

The photoelectrons are emitted from the cathode with negligible kinetic energy (Ki≈0K_i \approx 0). As they travel through a potential difference V=500 VV = 500 \text{ V} to the anode, they gain kinetic energy KK: K=eV=1.6×10−19 C×500 V=8×10−17 JK = e V = 1.6 \times 10^{-19} \text{ C} \times 500 \text{ V} = 8 \times 10^{-17} \text{ J}

The linear momentum pp of each photoelectron upon reaching the anode is given by: p=2meKp = \sqrt{2 m_e K}

Substituting the given mass of the electron me=9×10−31 kgm_e = 9 \times 10^{-31} \text{ kg}: p=2×(9×10−31 kg)×(8×10−17 J)=144×10−48=12×10−24 kg⋅m/sp = \sqrt{2 \times (9 \times 10^{-31} \text{ kg}) \times (8 \times 10^{-17} \text{ J})} = \sqrt{144 \times 10^{-48}} = 12 \times 10^{-24} \text{ kg}\cdot\text{m/s}

Step 3: Force experienced by the anode

Since all photoelectrons hit the anode normally and are fully absorbed, the rate of momentum transfer to the anode gives the average force FF: F=dNedt×pF = \frac{dN_e}{dt} \times p

F=(2×1020 s−1)×(12×10−24 kg⋅m/s)=24×10−4 NF = (2 \times 10^{20} \text{ s}^{-1}) \times (12 \times 10^{-24} \text{ kg}\cdot\text{m/s}) = 24 \times 10^{-4} \text{ N}

Comparing this with the given expression F=n×10−4 NF = n \times 10^{-4} \text{ N}, we get: n=24n = 24

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