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UV radiation with a 200 nm wavelength shines on a gold electrode in a photoelectric cell. What is the maximum kinetic energy, in eV, of photoelectrons ejected from the gold

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okpalawalter8

For a UV radiation with a 200 nm wavelength is mathematically given as the maximum kinetic energy, in eV, of photoelectrons ejected from the gold

K.E=1.779*10^{19}J

K.E=1.112ev

What is the maximum kinetic energy, in eV, of photoelectrons ejected from the gold?

Generally, the equation for the photo electric equation  is mathematically given as

[tex]hc/\lambda=K.E+\phi[/tex]

Therefore

K.E=(6.626*10^{-34}*3*10^8/2*10^{-7})-5.1*1.6*10^{-19}

K.E=1.779*10^{19}J

In conclusion, in eV, of photoelectrons ejected from the gold

K.E=1.112ev

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