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A narrow beam of monoenergetic electrons falls normally on the surface of a Ni single crystal. The reflection maximum of fourth order is observed in the direction forming an anle theta=55^(@) with the normal to the surface at the energy of the electrons equal to T= 180eV. Calcualte the corresponding value of the interplanar distance. |
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Answer» Solution :Path difference is `d+d cos theta= 2d "cos"^(2)(theta)/(2)` Thus for reflection MAXIMUM of the `K^(TH)` order `2 d "cos"^(2)(theta)/(2)k lambda=k(2pi ħ)/(sqrt(2mT))` Hence `d=(kpi ħ)/(sqrt(2mT))"sec"^(2)(theta)/(2)` Subsatitution with `k=4` gives `d= 0.232nm` 
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