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Two identical photons of energy $E (4 e V \leq E \leq 8 e V)$ fall on two different metals whose work functions are in the ratio of $1 : 2.$ The ratio of the kinetic energies of the most energetic electrons coming from each metal is $2 : 1.$ Work function of one of the metal is 4 eV. What is the energy of photon in eV.

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6 eV

5 eV

4 eV

8eV

answer is A.

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Detailed Solution

Let $ϕ_{1} = 4 e V,$ then $ϕ_{2} = 2 e V$
$(E - ϕ)$ represent kinetic energy of most energetic electron.
$E - ϕ_{2} = 2 (E - ϕ_{1})$ $\Rightarrow E = 6 e V$

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