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
Answers
Answer:
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
Explanation:
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Explanation:
K.E ( kinetic energy ) = h ( f - f₀ )
h ----- > plank's constant
f₀ ------> Threshold frequency
f --------> Frequency
------------------------------------------------
-------------------------------------------------------------------------
h = 6.626 × 10⁻³⁴
f₀ = 7× 10¹⁴ Hz
f = 1 × 10¹⁵ Hz
K.E = (6.626 ×10⁻³⁴ ) ₓ [ (1 × 10¹⁵) - (7× 10¹⁴)]
= 1.987 × 10⁻¹⁹ J //