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Q.

When a certain metal was irradiated with light of frequency 3.2 × 1016 Hz, the photoelectrons emitted had twice the kinetic energy as did photoelectrons when the same metal was irradiated with light of frequency 2.0 × 10¹⁶Hz. The threshold frequency for the metal is:

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a

1.2 × 10¹⁶ Hz

b

8 × 10¹⁵ Hz

c

8 × 10¹⁶ Hz

d

1.2 × 10¹⁵ Hz

answer is B.

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Detailed Solution

$KE = hν - {hν}_{o} given KE 2 = KE 1 v_{2} - v_{o} = \frac{{KE}_{2}}{h} v_{1} - v_{0} = \frac{{KE}_{1}}{h} \frac{v_{2} - v_{o}}{v_{1} - v_{0}} = \frac{{KE}_{2}}{{KE}_{1}} = \frac{2 {KE}_{1}}{{KE}_{1}} = 2 v_{2} - v = 2 \times v_{1} - v_{0} on solving v_{o} = 8.0 \times 10^{15} Hz$

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