When photon of energy 4.0 eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TA eV end de-Broglie wavelength $\lambda _A$. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50 eV is TB = (TA – 1.5) eV. If the de-Broglie wavelength of these photoelectrons $\lambda _B$ = 2$\lambda _A$, then the work function of metal B is :
Solution
We know, de-Broglie wavelength
<br><br>$\lambda$ = ${h \over p} = {h \over {\sqrt {2m{K_e}} }}$
<br><br>$\therefore$ $\lambda \propto {1 \over {\sqrt {{K_e}} }}$
<br><br>So ${{{\lambda _A}} \over {{\lambda _B}}} = \sqrt {{{{T_B}} \over {{T_A}}}}$
<br><br>$\Rightarrow$ ${\left( {{1 \over 2}} \right)^2}$ = ${{{{T_A} - 1.5} \over {{T_A}}}}$
<br><br>On solving T<sub>A</sub> = 2 eV
<br><br>$\therefore$ T<sub>B</sub> = T<sub>A</sub> - 1.5 = 0.5 eV
<br><br>Also T<sub>B</sub> = 4.5 - $\phi$<sub>B</sub>
<br><br>$\Rightarrow$ 0.5 = 4.5 - $\phi$<sub>B</sub>
<br><br>$\Rightarrow$ $\phi$<sub>B</sub> = 4 eV
About this question
Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: Photoelectric Effect
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