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Physics Question on Photoelectric Effect

When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TAT_A expressed in eV and de-Broglie wavelength λA\lambda_A. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is TB=(TA−1.50eV).T_B=(T_A-1.50eV). If the de-Broglie wavelength of these photoelectrons is λB=2λA,\lambda_B=2\lambda_A, then

A

the work function of A is 2.25 eV

B

the work function of B is 4.20 eV

C

TA=2.00eVT_A=2.00 eV

D

TB=2.75eVT_B=2.75 eV

Answer

TA=2.00eVT_A=2.00 eV

Explanation

Solution

Kmax=EWK_{max}=E_W Therefore, TA=4.25−WAT_A=4.25-W_A.....(i) TB(TA−1.50)=4.70−WBT_B(T_A-1.50)=4.70-W_B....(ii) From Eqs. (i) and (ii), WB−WA=1.95eVW_B-W_A=1.95 eV....(iii) de-Broglie wavelength is given by λ=h2Km\lambda=\frac{h}{\sqrt{2Km}} or λ∝1K\lambda \propto \frac{1}{\sqrt K} K = KE of electron ∴λBλA=KAKB\therefore \frac{\lambda_B}{\lambda_A}=\sqrt{\frac{K_A}{K_B}} or 2=TATA−1.52=\sqrt{\frac{T_A}{T_A-1.5}} This gives, TA=2eVT_A = 2 eV From E (i) WA=4.25−TA=2.25eVW_A=4.25-T_A=2.25 eV From E (iii) WB=WA+1.95eV=(2.25+1.95)eVW_B=W_A+1.95 eV=(2.25+1.95)eV or WB=4.20eVW_B=4.20 eV TB=4.70−WB=4.70−4.20T_B=4.70-W_B=4.70-4.20 =0.50 eV