Question

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?

Answer 1

By the law of conservation of momentum,

|P_C| + |P_A| + |P_B|

Let us first take case I when both P_A and P_B are positive,

then `λ_C = (λ_Aλ_B)/(λ_A + λ_B)`

In the second case when both P_A and P_B are negative,

then `λ_C = (λ_Aλ_B)/(λ_A + λ_B)`

In case III when P_A > 0, P_B < 0 i.e., P_A is positive and P_B 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 P_A < 0, P_B > 0, i.e., P_A, is negative and P_B 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