Question

In a photoelectric effect experiment, irradiation of a metal with light of frequency 5.2 x 10^14
yields electrons with maximum kinetic energy (KE) 1.3 x 10-19 J. Calculate the threshold
frequency (4) for the metal
13.24 - 100 g1 B) 2.14 × 10 5 9 1.24 × 104 51 D) 2.24 × 102 51
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Answer 1

Answer:

Solution :

We know that

hv=hv0+KE

or v0=v-

KE

h

KE=1.3×10-19J,v=5.2×1014s-1,h=6.626×10-34Js

:. Theshold frequency

v0=5.2×1024s-1-

1.3×1019J

6.626×10-34Js

=5.2×1014s-1-1.96×1014s-1

3.24×1014s-1

Answer 2

Answer:

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