Q.9
When a certain metal was irradiated with light having a frequency of 3.0 x 1016 sec !, the photoelectrons
emitted had twice the kinetic energy as did photoelectrons emitted when the same metal was irradiated
with light having a frequency of 2.0 x 10 sec '. Calculate v, for the metal.
Answers
Answered by
1
Answer:
Explanation:
ANSWER
Applying photoelectric equation,
KE=hv−hv
0
or (v−v
0
)=
h
KE
Given, KE
2
=2KE
1
v
2
−v
0
=
h
KE
2
...(i)
and v
1
−v
0
=
h
KE
1
...(ii)
Dividing equation (i) by equation (ii),
v
1
−v
0
v
2
−v
0
=
KE
1
KE
2
=
KE
1
2KE
1
=2
or v
2
−v
0
=2v
1
−2v
0
or v
0
=2v
1
−v
2
=2(2.0×10
16
)−(3.2×10
16
)
=8.0×10
15
Hz.
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