The kinetic energy of the electron emitted when light of frequency 3.5 × 1015 Hz falls on a metal surface having threshold frequency 1.5 × 1015 Hz is h=6.6 × 10−34 Js

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1.32 × 10−18 J

3.3 × 10−18 J

6.6 × 10−19 J

1.98 × 10−19 J

answer is A.

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Detailed Solution

KE=hv−v0=6.6 × 10−34 3.5 × 1015−1.5 × 1015= 1.32 × 10−18 J

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