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 $\lambda_A$. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is $T_B=(T_A-1.50eV).$ If the de-Broglie wavelength of these photoelectrons is $\lambda_B=2\lambda_A,$ then

• A. the work function of A is 2.25 eV
• B. the work function of B is 4.20 eV
• C. $T_A=2.00 eV$
• D. $T_B=2.75 eV$

Answer 1

The Correct Option is C

$K_{max}=E_W$ Therefore, $T_A=4.25-W_A$.....(i) $T_B(T_A-1.50)=4.70-W_B$....(ii) From Eqs. (i) and (ii), $W_B-W_A=1.95 eV$....(iii) de-Broglie wavelength is given by $\lambda=\frac{h}{\sqrt{2Km}}$ or $\lambda \propto \frac{1}{\sqrt K}$ K = KE of electron $\therefore \frac{\lambda_B}{\lambda_A}=\sqrt{\frac{K_A}{K_B}}$ or $2=\sqrt{\frac{T_A}{T_A-1.5}}$ This gives, $T_A = 2 eV$ From E (i) $W_A=4.25-T_A=2.25 eV$ From E (iii) $W_B=W_A+1.95 eV=(2.25+1.95)eV$ or $W_B=4.20 eV$ $T_B=4.70-W_B=4.70-4.20$ =0.50 eV