When photons of energy 4 25 ev strike the surface of a metal A The ejected photoelectrons have maximum - Chemistry - Atomic Structure

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

Vartika Jain · Accepted Answer

Dear Student,

λ=hmv=hpK E
 , T = p22mTherefore, TBTA=(pBpA)2=(λAλB)2=14Therefore, TA=4TBBut, TB=(TA-1
5)eVSo, TB=0
5 eV and TA=2 eVSince work function of a metal is equal to energy of incident minus maximum kinetic energy of photoelectrons, thereforeWork function of A = (4
25-2)=2 25 eVWok function of B=((4 2-0 5)=3
7 eVHence, the correct answer is (D)