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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-hνo given KE2=KE1 v2-vo=KE2h v1-v0=KE1h v2-vov1-v0=KE2KE1=2KE1KE1=2 v2-v= 2×v1-v0 on solving  vo=8.0 ×1015Hz

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