 Quote by Reshma
a] A beam of light of wavelength 400nm and power 1.55mW is directed at the cathode of a photoelectric cell. If only 10% of the incident photons effectively produce photoelectrons, find the current due to these electrons.
b]If the wavelength of light is now reduced to 200nm, keeping the power same, the kinetic energy of the electrons is found to increase by a factor of 5. What are the values of the stopping potentials for the two wavelengths?
My Work:
a]Given the wavelength of incident light I calculated the incident energy as [itex]h\nu = h{c\over \lambda} = 3.1 eV[/itex]
The doubt in my mind is over the term "effective". If 10% of the incident photons is utilized for producing photoelectrons, then it should be equal to the Kinetic energy of the electrons (KE). Hence, [itex]KE = 0.1 h\nu[/itex] in eV.
Power = Voltage x Current
Current(I) = 1.55mW/0.31V
Please correct me if I am wrong.
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Yes, this not correct.
You have the incident power and the energy of each photon. From this, you can calculate the incident photon rate striking the metal. Assuming 10% of the photons incident on the metal produce photoelectrons, you can find the rate of electron emission, and hence the current produced.
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b] The stopping potential [itex]V_s[/itex] is related to the kinetic energy by [itex]KE_{max} = eV_s[/itex] and we have the relationship:
[tex]{h\over e} = \frac{V_{S2} - V_{S1}}{\nu_2 - \nu_1}[/tex]
I am a little confused over whether I should use this relation or simply add 5 to the KE determined from the previous exercise and divide it by e to obtain the stopping potential.
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Write down the complete equation for the PE effect (twice - once for 400nm and again for 200nm). Use the fact that the KE in the second case is 5 times the KE in the first case.
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