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प्रश्न
A neutron beam of energy E scatters from atoms on a surface with a spacing d = 0.1 nm. The first maximum of intensity in the reflected beam occurs at θ = 30°. What is the kinetic energy E of the beam in eV?
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उत्तर
By the law of conservation of momentum,
|PC| + |PA| + |PB|
Let us first take case I when both PA and PB are positive,
then `λ_C = (λ_Aλ_B)/(λ_A + λ_B)`
In the second case when both PA and PB are negative,
then `λ_C = (λ_Aλ_B)/(λ_A + λ_B)`
In case III when PA > 0, PB < 0 i.e., PA is positive and PB is negative,
`h/λ_C = h/λ_A - h/λ_B = ((λ_B - λ_A)h)/(λ_Aλ_B)`
⇒ `λ_C = (λ_Aλ_B)/(λ_B - λ_A)`
And In case IV when PA < 0, PB > 0, i.e., PA, is negative and PB is positive.
∴ `h/λ_C = h/λ_A + h/λ_B`
⇒ `((λ_A - λ_B)h)/(λ_Aλ_B)`
⇒ `λ_C = (λ_Aλ_B)/(λ_A - λ_B)`
Now, KE = `1/2 mv^2 = 1/2 (m^2v^2)/m = 1/2 P^2/m`
= `1/2 xx (6.62 xx 10^-24)^2/(1.67 xx 10^-27)`
= 0.21 eV
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