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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? |
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Answer» Solution :Here ,if perpendicular distance between two consecutitive atomic layers is d then the path difference betweenthe 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` |
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