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Q.

When a certain metal was irradiated with light of frequency 3.2×1016Hz , 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

1.2×1016Hz

b

8×1015Hz

c

8×106Hz

d

1.2×1015Hz

answer is B.

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Detailed Solution

hϑ1=hϑ0+KE1 hϑ2=hϑ0+KE2 h(ϑ1−ϑ0)h(ϑ2−ϑ0)=E1E2 3.2×1016−ϑ02.0×1016−ϑ0=21 3.2×1016−ϑ0=2(2.0×1016−ϑ0) 3.2×1016−ϑ0=4×1016−2ϑ0                       ϑ0=0.8×1016                      ϑ0=8×1015
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