derive the equation of de brogile wave length of a proton of potential V
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Explanation:
As de-Broglie wavelength is given by λ=
mv
h
=
2m(K.E)
h
λ=
2mqV
h
where v= accelerating potential
q=charge
λ
p
=
2m
p
q
p
v
h
α=
4
He
++
⇒rq
α
=2q
p
m
α
=4m
p
λ
α
at same accelerating potential
λ
α
=
2×4m
p
×2q
p
×v
h
λ
α
=
2m
p
q
p
v
×
8
h
=
2
2
λp
Given λ
p
=λ
⇒λ
α
=
2
2
λ
.
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