1.5 mW of 400 nm light is directed at a photoelectric cell. If 0.1% of the incident photons produce photoelectrons, the current in the cell is:
Correct answer: A. 0.48µA
- A. 0.48µA
- B. 0.42mA
- C. 0.48 mA
- D. 0.42µA
Explanation
For wavelength 400 nm, energy of each photon will be:E = hc/λn number of photon will have energyne = 1.5mW0.1% photons extract electronsNow, nq = nenq = 1.5 x 1/100= 0.48µAThis is the following solution:
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