The frequency and work function of an incident photon are
v and Φ₀. If v₀ is the threshold frequency then necessary
condition for the emission of photoelectron is
(a) v < v₀ (b) v = v₀/2
(c) v ≥ v₀ (d) None of these
Answers
Answered by
0
Answer:
OPTION C IS CORRECT
Explanation:
Energy of incident radiation E=hν
Work function ϕ=hν
0
For ejection of photoelectron
E≥ϕ
or
ν≥ν
0
Answered by
0
Answer:
The Einstein's equation for photoelectric effect is given by , KE=hν−hν
0
where KE=maximum kinetic energy of electron, h= Planck's constant , ν
0
= threshold frequency and ν= frequency of incident light .
Also KE=
2
1
mv
2
where v= maximum velocity of electron.
In first case,
2
1
mv
1
2
=h(2ν
0
)−hν
0
=hν
0
....(1) and
In second case,
2
1
mv
2
2
=h(5ν
0
)−hν
0
=4hν
0
....(2)
From (1) and (2),
2
1
mv
2
2
=4×
2
1
mv
1
2
or v
2
=2v
1
=2×(4×10
6
)=8×10
6
m/s
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