Question

Directions: The following question has four choices, out of which one or more are correct.When photons of energy 4.25 eV strike the surface of a metal $A$, the ejected photoelectrons have minimum kinetic energy $T_A$ expressed in eV and de Broglie wavelength ${\lambda }_A.$ The maximum kinetic energy ofphotoelectrons 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) TA = 2.00 eV
• (D) TB = 2.75 eV

Answer 1

The Correct Option is A, D

Explanation:
$K_{max}=E-W$Therefore,$T_A=4.25-W_A\dots$ (i)$T_B=(T_A-1.50)=4.70-W_B\dots$ (ii)From Eqs. (i) and (ii),$W_B-W_A=1.95eV\dots$ (iii)de Broglie wavelength is given by$\lambda =\frac{h}{\sqrt{2Km}}$ or $\lambda \propto \frac{1}{\sqrt{K}}$$\frac{{\lambda }_A}{{\lambda }_B}\propto \sqrt{\frac{K_B}{K_A}}$$2=\sqrt{\frac{T_A}{T_A-1.5}}$$$\begin{array}{l}\text{ This gives }{T}_A=2eV\\ \text{ From equation (i), }{W}_A=4.25-T_A=2.25eV\\ \text{ From equation (iii), }{W}_B=W_A+1.95eV=(2.25+1.95)eV\\ \text{ Or, }{W}_B=4.20eV\\ T_B=4.70-W_B=4.70-4.20\\ =0.50eV\end{array}$$