Question

When photons of energy 4.25 ev strike the surface of a metal A. The ejected photoelectrons have maximum kinetic energy (T(A) (expressed in ev) and de-Broglie wavelength (NA). The max kinetic energy of photoelectrons liberated from another Metal B by photons of enegy 4.2 eV is TB.Where TB (TA-1.5). If De-Broglie wave length of these photoelectrons AB (nB A), then which of the following is not correct (A) The work function of A is 2.25 eV. (B) The work function of B is 3.7 eV (C) TA=2.0 eV (D) TB=0.75 eV

Answer 1

Option D is incorrect

This is a very difficult sum and it took me nearly 17 mins to solve.

Kindly take a look at the steps in the given attachment pic.