When a certain metal was irradiated with light of frequency
3.2 x 1016 s-1, the photoelectrons emitted had twice the
kinetic energy as did the photoelectrons emitted when
the same metal was irradiated with light of frequency
2.0x1016 s-1. Calculate threshold frequency for the metal.
[Hint. K.E. = hv -hvo; hví - hvo = 2 (hv 2 - hvo);
or Vo = 2v2 - Vi]
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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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