When the momentum of a proton is changed by an amount p0,the corresponding change in the de-broglie wavelength is found to be 0.25 percent.then the oroginal momentum of the proton was?
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We have λ=
p
h
λ∝
p
1
⇒d
p
Δp
=
λ
Δλ
⇒
∣
∣
∣
∣
∣
p
Δp
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
λ
Δλ
∣
∣
∣
∣
∣
p
p
0
=
100
0.25
p
p
0
=
400
1
p=400p
0
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