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Q.
Light of wavelength $4000\, \mathring{A}$ is incident on a metal plate whose work function is $2 \,eV$. The maximum kinetic energy of the emitted photo-electron will be :
BHUBHU 2002
Solution:
When maximum kinetic energy of photoelectrons is $E_{K}$, and $W$ is work function of metal, then from Einstein's photoelectric equation.
$E_{K}=h v-W$
where $h v$ is energy of photon absorbed in the metal.
Also $ E =h v=\frac{h c}{\lambda} $
$ E =\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{4000 \times 10^{-10} \times 1.6 \times 10^{-19}} eV $
$=0.31 \,eV $
$ \therefore E_{k} =3.1-2 $
$=1.1\, eV$
Note : Photoelectrons are emitted when energy of incident photon is more than work function of the metal