# Help on Photoelectric Effect problem

1. Oct 22, 2007

### Reshma

Help on "Photoelectric Effect" problem

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 $h\nu = h{c\over \lambda} = 3.1 eV$
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, $KE = 0.1 h\nu$ in eV.
Power = Voltage x Current
Current(I) = 1.55mW/0.31V

Please correct me if I am wrong.

b] The stopping potential $V_s$ is related to the kinetic energy by $KE_{max} = eV_s$ and we have the relationship:
$${h\over e} = \frac{V_{S2} - V_{S1}}{\nu_2 - \nu_1}$$
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.

2. Oct 22, 2007

### Gokul43201

Staff Emeritus
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.

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.

3. Oct 26, 2007

### Reshma

The incident Power is 1.55mW and energy of each photon (400nm wavelength) is 3.1eV. What is the incident photon rate? Isn't that equal to the power?
Thanks, I got this! It was just a matter of rearranging the terms.