The threshold wavelength is 2000 \[{A^0}\], the work function is:
(A) 6.25 eV
(B) 6.2 eV
(C) 6.2 MeV
(D) 6.2 KeV

Hint Work function for a particular metal is the minimum energy a photon must possess in order to eject an electron from the surface of the metal on which the light is incident. The threshold wavelength is the maximum wavelength of the photon that can eject an electron. The relation between work function and threshold wavelength of the photon was given by Einstein and we just have to substitute the values in them.

Complete step-by-step solution
Einstein gave the work function of a metal as
 \[\phi = \dfrac{{hc}}{\lambda }\]
Here
 \[\phi \] is the work function
h is the planck's constant
c is the speed of light in vacuum
 \[\lambda \] is the wavelength of the light source incident on the surface
We know the values of h and c, and \[\lambda \] is given in the question, substituting the values of given we get,
 \[\phi = \dfrac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{2000 \times {{10}^{ - 10}}}} = 9.9939 \times {10^{ - 19}}\]
This value is in terms of J, but we need to find a value in eV, so we will divide the answer with \[1.6 \times {10^{ - 19}}\] ,
 \[\phi = \dfrac{{9.9939 \times {{10}^{ - 19}}}}{{1.6 \times {{10}^{ - 19}}}}\]
=6.2 eV

Therefore the correct answer is option B.

Note The value of the product of planck's constant with the speed of light in terms of eV is written as 1240eV-nm. This can be obtained by converting m to mm which will give the required quantity in J. To convert it into eV, you need to divide it by the charge of an electron which is \[1.6 \times {10^{ - 19}}\] C.