The correct option is C 3×10−20 J
Given : wavelength of the incident radiation (λ)=2210 nm
Also, the minimum energy required by the electron to overcome the attractive forces ( E0) =0.6×10−19 J
Finding the energy of the incident radiation (E)
We know that, E=hcλ
Where, h= Planck's constant =6.626×10−34 Js
c= speed of light =3×108 m/s
λ= Wavelength of the incident radiation
=2210 nm
=2210×10−9 m
∴E=6.626×10−34 J.s×3×108 m/s2210×10−9 m
E=(6626×10−32210×10−9×10−34×3×108) J
=9×10−20 J
E=0.9×10−19 J
According to Einstein's photoelectric equation,
E=E0+K.Emax
Where, E is the energy of the incident light
E0 is the work function.
KEmax is the maximum kinetic energy of the ejected electron.
Substituting the values of E and E0,
0.9×10−19 J=0.6×10−19 J+K.Emax
∴K.Emax=(0.9−0.6)10−19 J
=0.3×10−19 J
K.Emax=3×10−20 J