photoelectric emission is observed from a metal surface for the frequencies v1 and v2 radiation. if the maximum kinetic energy of the two electrons are in the ratio 1:2 then how is threshold frequency expressed
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Answer:
hv
1
=hv
0
+
2
1
mu
1
2
..(1)
hv
2
=hv
0
+
2
1
mu
2
2
.(2)
2
1
mu
1
2
=(
k
1
)
2
1
mu
2
2
From equation (1)-
hv
1
=hv
0
+
2k
1
mu
2
2
3)
2
1
mu
2
2
=khv
1
−khv
0
(4)
From equations (2) and (4), we get-
hv
2
=hv
0
−khv
0
+khv
1
v
0
(1−k)=v
2
−kv
1
v
0
=
(k−1)
kv
1
−v
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