please answer this question and explain how to do it The metal forming cathode u

Question

please answer this question and explain how to do it

The metal forming cathode used in a certain Photoelectric Effect experiment (not ours) has a work function of 4.00 eV. What is the maximum speed of the electrons that are liberated by photons possessing a frequency of 3.0 Times 1015 Hz? In this case, the photoelectrons are non-relativistic (i.e. they are travelling very slow compared to the speed of light in vacuum), so one can use the Newton form of the kinetic energy in this calculation, namely, KE = 1/2 me upsilon 2.

Explanation / Answer

Energy of photons = hv = 6.6*10^-34*3*10^15 = 19.8*10^-19;;;;

K.E max= hv - W = 19.8*10^-19 -[ 4*1.6*10^-19] = 13.4*10^-19 = 0.5mv^2

=> v max= 1.7*10^6 m/s

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