The work functions of two metals (M A and M B ) are in the 1 : 2 ratio. When these metals are exposed to photons of energy 6 eV, the kinetic energy of liberated electrons of M A : M B is in the ratio of 2.642 : 1. The work functions (in eV) of M A and M B are respectively. [JEE Main 2026, 23 Jan (Shift 2)]

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YESPYQ · Accepted Answer

Correct answer: c) 2.3, 4.6. $\begin{matrix}{\mathrm{KE}}_{\max }=\mathrm{E}−\mathrm{ϕ} \ {\left({\mathrm{KE}}_{\max }\right)}_{1}=6−{\mathrm{ϕ}}_{1}\text{ }.....\text{ }(1) \ {\left({\mathrm{KE}}_{\max }\right)}_{2}=6−{\mathrm{ϕ}}_{2}\text{ }.......(2)\end{matrix}$ By eq. (1) divide eq. (2) $\frac{{\left(K{E}_{\max }\right)}_{1}}{{\left(K{E}_{\max }\right)}_{2}}=\frac{2.642}{1}=\frac{6−{ϕ}_{1}}{6−{ϕ}_{2}}$ $\frac{2.642}{1}=\frac{6−{\mathrm{ϕ}}_{1}}{6−2{\mathrm{ϕ}}_{1}}$ $\begin{matrix}{\mathrm{ϕ}}_{1}=2.3\mathrm{eV} \ {\mathrm{ϕ}}_{2}=4.6\mathrm{eV}.\end{matrix}$

The work functions of two metals (M A and M B ) are in the 1 : 2 ratio. When these metals are exposed to photons of ener…

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