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Q.

The work functions of metals A and B are in the ratio 1 : 2. If light of frequencies $f$ and $2 f$ are incident on metal surfaces of A and B respectively, the ratio of the maximum kinetic energies of photo electrons emitted is (f is greater then threshold frequency of A, 2f is greater than threshold frequency of B

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a

1 : 3

b

1 : 2

c

1 : 1

d

1 : 4

answer is B.

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Detailed Solution

$h v = w + k E$

$k E = h v - w$

$k E = h f - w$

$k E_{2} = 2 h f - 2 w$

$= 2 (h f - w)$

$\frac{k E_{1}}{k E_{2}} = \frac{1}{2}$

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