The work functions of two metals A and B are in the ratio $1 : 3$ . If light of frequency f and $3 f$ are incident on the surface of A and B respectively, then the ratio of maximum kinetic energies of photoelectrons is?

1 : 2

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1 : 3

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2 : 1

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3 : 1

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Solution

The correct option is C $1 : 3$
$K E_{m a x} = h v - ϕ$
For metal A,
$K_{A} = h f = ϕ_{A}$ .........(i)
For metal B,
$K_{B} = 3 h f - ϕ_{B}$ .............(ii)
$∴$ From Eqs. (i) and (ii), we get
$\frac{K_{A}}{K_{B}} = \frac{h f - ϕ_{A}}{3 h f - ϕ_{B}}$
$= \frac{(\frac{h f}{ϕ_{A}} - 1)}{\frac{3 h f}{ϕ_{A}} - \frac{ϕ_{B}}{ϕ_{A}}} = \frac{(\frac{h f}{ϕ_{A}} - 1)}{\frac{3 h f}{ϕ_{A}} - \frac{3}{1}}$
$= \frac{(\frac{h f}{ϕ_{A}} - 1)}{3 (\frac{h f}{ϕ_{A}} - 1)} = \frac{1}{3}$
$∴ K_{A} : K_{B} = 1 : 3$ .

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