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For what kinetic energy of a neutron, will the associated de-Broglie wavelength be 1.32xx10^(-10)m ? Given that mass of a neutron=1.675xx10^(-27)kg. |
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Answer» SOLUTION :It is GIVEN that `lamda=1.32xx10^(-10)m and m_(n)=1.675xx10^(-27)kg` `because lamda=(h)/(sqrt(2m_(n)K)) implies K=(h^(2))/(2m_(n)lamda^(2))`. `THEREFORE`Kinetic ENERGY of neutron `K=((6.63xx10^(-34))^(2))/(2xx(1.675xx10^(-27))xx(1.32xx10^(-10))^(2))=7.53xx10^(-21)`. |
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