A neutron beam of energy 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 `theta=30^(@)` .What is the kinetic energy E of the beam in eV?

Explanation

Here ,if perpendicular distance between two consecutitive atomic layers is d then the path difference between the rays reflected at angle `theta` is `2dsintheta` which should be equal to `nlambda` for constructive interference in the present case.
Thus.
`2dsintheta=nlambda`
`therefore 2sin30^(@)=(1)lambda`
`therefore lambda`=d=0.1nm.........(1)
Here ,reflection of neutron is given like a wave and so its momentum would be `p=(h)/(lambda)` and its energy would be,
`E=(p^(2))/(2m)`
`therefore E=(h^(2))/(2mlambda^(2))`
`therefore E=((6.625xx10^(-34))^(2))/((2)(1.67xx10^(-27))(0.1xx10^(-9))^(2))`
`therefore E=13.14xx10^(-68+27+20)J`
`therefore E=13.14xx10^(-21)J`
`therefore E=(13.14xx10^(-21))/(1.6xx10^(-19))eV`
`therefore E=8.2125xx10^(-2)eV`