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A neutron beam of energy E scatters from...

A neutron beam of energy E scatters from atoms on a surface with a spacing d=0.1nm. The first maximum of intensity in the reflected beam occurs at `theta=30^(@)`. What is the kinetic energy of E of the beam in eV?

Text Solution

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Give,
Atomic spacing , d = 0.1 nm,
`theta=30^@`
Order of maxima , n = 1
Using Bragg.s law, we have
Using Bragg.s law , we have
`2d sin theta = nlamda`
`implies2xx0.1xxsin 30^@ = 1 xx lamda`
`implies lamda = 0.1 nm = 10^(-10)m`
Also, momentum of particle is
`p=h/lamda=(6.63xx10^(-34)Js)/(10^(-10)m)`
`=6.63xx10^(-24)"kg m s"^(-1)`
Kinetic energy of beam is
`E=p^2/(2m)=(1/2)xx((6.63xx10^(-24)))/((1.67xx10^(-27)))J`
`impliesE=(1/2)xx((6.63xx10^(-24))^2)/((1.07xx10^(-27))xx(1.6xx10^(-19)))eV`
= 0.082 eV
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