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Q. Light of wavelength $ \lambda = 400 \, nm$ is incident on a photosensitive metal whose work function is $2 eV$. The maximum kinetic energy of photo electrons emitted will be

COMEDKCOMEDK 2008Dual Nature of Radiation and Matter

Solution:

According to Einstein's photoelectric equation, $ K = \frac{hc}{\lambda} - \phi $
Here, $ \lambda = 400 \, nm , hc = 1242 \, eV - nm$
$ \phi = 2 \, eV$
$ \therefore \, K = \frac{1242 eV -nm}{400 nm} - 2eV $
$ \, \, \, \, \, \, \, \, = 3.1 eV - 2 eV = 1.1 eV $