The minimum frequency of light that will cause photoelectric emission is 5.25*10^14Hz , when the surface is eliminated with another radiation the electrons leave with a speed of 7.25*10^5m/s find the frequency of the second
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ANSWER:
STEP BY STEP EXPLANATION:
v=8.41×10
14
Hz
V
max
=7.5×10
5
ms
−1
h=6.625×10
−34
Js
−1
m=9.1×10
−31
kg
hv=ϕ
0
+K
max
hv=hv
0
+
2
1
mV
max
2
hv−hv
0
=
2
1
mV
max
2
(v−v
0
)=
2h
1
mV
max
2
=
2×6.625×10
−34
1
×9.1×10
−31
×(7.5×10
5
)
2
(v−v
0
)=3.8632×10
14
v
0
=8.41×10
14
−3.8632×10
14
=4.5468×10
14
Hz
ϕ
0
=hv
0
=6.625×10
−34
×4.5468×10
14
=3.01226×10
−19
ϕ
0
=
1.6×10
−19
3.01226×10
−19
=1.8826eV
Mark me as brainlist :-)
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