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Question: A neutron beam of energy K scatters from atoms on a surface with a spacing d = 0.1 nm. The first max...

A neutron beam of energy K 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=300\theta = 30^{0}. The kinetic energy K of the beam in eV is :

A

0.15

B

2.13

C

1.21

D

0.08

Answer

0.08

Explanation

Solution

: Here ,d=0.1nm,θ=30∘,n=1d = 0.1nm,\theta = 30{^\circ},n = 1

According to Bragg’s law

2dsin⁡θ=nλ2d\sin\theta = n\lambda

2×0.1×sin⁡30∘=1×λ2 \times 0.1 \times \sin 30{^\circ} = 1 \times \lambda

Or λ=0.10nm=10−10m\lambda = 0.10nm = 10^{- 10}m

As p=hλ=6.63×10−3410−10=6.63×10−24kgms−1p = \frac{h}{\lambda} = \frac{6.63 \times 10^{- 34}}{10^{- 10}} = 6.63 \times 10^{- 24}kgms^{- 1}

Kinetic energy,

}{= \frac{1}{2} \times \frac{(6.63 \times 10^{- 24})^{2}}{(1.67 \times 10^{- 27})(1.6 \times 10^{- 19})}eV = 0.08eV}$$