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

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)

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