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Question: In a photoelectric experiment, photo-efficiency is 1%. At anode, electric potential is twice the mag...

In a photoelectric experiment, photo-efficiency is 1%. At anode, electric potential is twice the magnitude of stopping potential which enables 50% of the emitted electrons to reach anode constituting photo-current. A monochromatic source of power 1 mW is used which emits photons of wavelength 100 nm. Find the value of photo-current. [Take hc = 1240 nm eV and e = 1.6 × 10^{-19} C]

Answer

4.032 × 10^{-7} A

Explanation

Solution

Explanation of the Solution

  1. Photon Energy Calculation: The energy of a single photon is calculated using Ephoton=hcλE_{photon} = \frac{hc}{\lambda}.
  2. Number of Photons Incident per Second: The power of the source is given by P=Nphoton×EphotonP = N_{photon} \times E_{photon}, allowing us to find the number of photons incident per second (NphotonN_{photon}).
  3. Number of Electrons Emitted per Second: The photo-efficiency (η\eta) relates the number of emitted electrons (Nelectrons_emittedN_{electrons\_emitted}) to the number of incident photons: Nelectrons_emitted=η×NphotonN_{electrons\_emitted} = \eta \times N_{photon}.
  4. Number of Electrons Collected per Second: The problem states that at the given anode potential, 50% of the emitted electrons reach the anode. This directly implies that the number of electrons collected per second (NcollectedN_{collected}) is 50% of the number of electrons emitted per second. The information about the anode potential being twice the stopping potential, while seemingly contradictory in a literal interpretation, is superseded by the explicit statement that 50% of electrons are collected, which defines the photocurrent.
  5. Photo-current Calculation: The photo-current (IphI_{ph}) is the rate of charge flow, calculated as Iph=Ncollected×eI_{ph} = N_{collected} \times e.

Calculations:

  • Photon Energy: Ephoton=hcλ=1240 nm eV100 nm=12.4 eVE_{photon} = \frac{hc}{\lambda} = \frac{1240 \text{ nm eV}}{100 \text{ nm}} = 12.4 \text{ eV} Convert to Joules: Ephoton=12.4 eV×(1.6×10−19 J/eV)=19.84×10−19 JE_{photon} = 12.4 \text{ eV} \times (1.6 \times 10^{-19} \text{ J/eV}) = 19.84 \times 10^{-19} \text{ J}

  • Number of Photons per Second: P=1 mW=1×10−3 W=1×10−3 J/sP = 1 \text{ mW} = 1 \times 10^{-3} \text{ W} = 1 \times 10^{-3} \text{ J/s} Nphoton=PEphoton=1×10−3 J/s19.84×10−19 J≈5.04×1014 s−1N_{photon} = \frac{P}{E_{photon}} = \frac{1 \times 10^{-3} \text{ J/s}}{19.84 \times 10^{-19} \text{ J}} \approx 5.04 \times 10^{14} \text{ s}^{-1}

  • Number of Electrons Emitted per Second: η=1%=0.01\eta = 1\% = 0.01 Nelectrons_emitted=η×Nphoton=0.01×(5.04×1014 s−1)=5.04×1012 s−1N_{electrons\_emitted} = \eta \times N_{photon} = 0.01 \times (5.04 \times 10^{14} \text{ s}^{-1}) = 5.04 \times 10^{12} \text{ s}^{-1}

  • Number of Electrons Collected per Second: Fraction collected = 50% = 0.50 Ncollected=0.50×Nelectrons_emitted=0.50×(5.04×1012 s−1)=2.52×1012 s−1N_{collected} = 0.50 \times N_{electrons\_emitted} = 0.50 \times (5.04 \times 10^{12} \text{ s}^{-1}) = 2.52 \times 10^{12} \text{ s}^{-1}

  • Photo-current: Iph=Ncollected×e=(2.52×1012 s−1)×(1.6×10−19 C)I_{ph} = N_{collected} \times e = (2.52 \times 10^{12} \text{ s}^{-1}) \times (1.6 \times 10^{-19} \text{ C}) Iph=4.032×10−7 AI_{ph} = 4.032 \times 10^{-7} \text{ A} Iph=0.4032 μAI_{ph} = 0.4032 \, \mu\text{A}