the threshold frequency for a metal is 7 10^14 s^-1 calculate the kinetic energy of an electron emitted when radiation of frequency v= 8* 10^15 s-1 hits the metal
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Answer:
Explanation:
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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