The threshold frequency vo for a metal is 8 x 10 to the power 14 per second What is the kinetic energy of an electron
emitted when metal is exposed to light having frequency v = 1.0 ×10 to the power 15 per second
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Answered by
2
Answer:
Given that,
v
o
=7.0×10
14
s
−1
v=1.0×10
15
s
−1
As we know, hv=hv
0
+
2
1
mv
2
i.e. h(v−v
0
)=K.E.
∴K.E.=(6.626×10
−34
)×(1.0×10
15
−7.0×10
14
)=6.626×10
−34
×10
14
(10−7)
∴K.E=19.878×10
−20
J
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0
Explanation:
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