The work function of metal is 1 eV. Light of
wavelength 3000 A is incident on this metal
surface. The maximum velocity of emitted
photoelectron will be
(1) 10 m s-1
(2) 1x103 m s-1
(3) 1x104m 5-1
(4) 1*106 m
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Answer:
We know that,
E=hv=
λ
h
c
=
3×10
−3
6.6×10
−34
×3×10
8
E=6.6×10
−19
J
we know that,
E=ϕ+KE
6.6×10
−19
−1.6×10
−19
=
2
1
mV
2
[1eV−16×10
−19
]
V
2
=
9.1×10
−31
5×10
−19
×2.1.1×10
12
V=1×10
6
m/s
Explanation:
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