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Photo electrons are emitted from a surface with a speed of 7×10 5 ms −1 . If the frequency of the incidence radiation is 8×10 14 Hz, the threshold frequency for the surface is |
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Answer» Answer: v=8.41×10 14 Hz V max
=7.5×10 5 −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 Explanation: Hope it helps you don't forget to follow and mark me as brainliest❣️ |
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