Question

Problem: Light strikes a sodium surface, causing photoelectric emission. The stopping potential for the ejected electrons is 5.0V, and the work function of sodium is 2.2eV. What is the wavelength of the incident light? Give your answer in nanometers.​

Answer 1

In photoelectric emission, KE of electrons is given by:

KE_max = E - phi

KE_max = h*c/lambda - phi

phi = work-function = 2.2 eV

max KE = e*V0

where V0 = stopping potential = 5 V

max KE = e*5 V = 5 eV

So,

lambda = hc/(KE_max + phi)

lambda = 6.626*10^-34*3*10^8/(5 eV + 2.2 eV)

lambda = 6.626*10^-34*3*10^8/(7.2 eV)

lambda = 6.626*10^-34*3*10^8/(7.2*1.6*10^-19)

lambda = 1.725*10^-7 m = 172.5*10^-9 m

lambda = wavelength of incident light = 172.5 nm

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