Question

When light of frequency 3.2 x 10to the power 16 Hz is used to irradiate a metal surface the maximum kinetic energy
of the emitted photoelectron is 3/4 of the energy of irradiating photon then the threshold frequency of the
metal would be:
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Answer 1

Answer:

I hope you understand........

Explanation:

Where work function = hνo

νo = threshold frequency

KEmax=43E

E=hν=6.62×10−34×3.2×1016=21.184×10−18J

hνo=E−0.75E=0.25E=0.25×21.184×10−18

⇒νo=8×1015Hz