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(a) For what kinetic energy of a neutron will the associated de-broglie wavelength be 1.40xx10^(-10)m? (b) Also find the de-Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of 3/2kT at 300K. |
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Answer» Solution :(a) Mass of neutron `m_(N)=1.675xx10^(-27)KG` and de-Broglie wavelength `lamda=1.40xx10^(-10)m` `because lamda=(h)/(p)=(h)/(sqrt(2m_(n)K))` `K=(h^(2))/(2m_(n)lamda^(2))=((6.63xx10^(-34))^(2))/(2xx(1.675xx10^(-27))xx(1.40xx10^(-10))^(2))=6.686xx10^(-21)J` `=(6.686xx10^(-21))/(1.6xx10^(-19))eV=4.174xx10^(-2)eV`. (b) As kinetic energy of neutron `K=(3)/(2)k_(B)T`, where `k_(B)=`Boltzmann.s constant`=1.38xx10^(-23)JK^(-1)` and T=300K (given) `therefore lamda=(h)/(sqrt(2m_(n)K))=(h)/(sqrt(3m_(n)k_(B)T))=(6.63xx10^(-34))/(sqrt(3xx(1.675xx10^(-27))xx(1.38xx10^(-23))xx300))=1.45xx10^(-10)m`. `=0.145nm`. |
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