The threshold frequency for a metal is 7.0 x 1014 s-1. Calculate the kinetic energy of an emitted electron
when radiation of frequency (ν) 1.0 x 1015 s
-1 hits the metal.
Answers
Answered by
2
Answer:
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
Answered by
0
Answer:
yes i knoe thw ans
KE = hv-hv0
7.0-1
6
Explanation:
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