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For what kinetic energy of a neutron will the associated de-Broglie wavelength be 1.40xx10^(-10)m? |
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Answer» Solution :`lambda=1.40xx10^(-10)m`, `H=6.63xx10^(-34) Js` `m=1.67xx10^(-27)`kg (a) de-Broglie wavelength of neutron, `lambda=(h)/(sqrt(2mK))` `therefore K=(h^(2))/(2mlambda^(2))` `=((6.63xx10^(-34))^(2))/(2xx1.67xx10^(-27)XX(1.4xx10^(-10))^(2))` `therefore K=6.686xx10^(-21)`J `=(6.686xx10^(-21))/(1.6xx10^(-19))eV` `therefore K=4.174xx10^(-2)eV` (b)Average kinetic energy of neutron Here `k_(B)=1.38xx10^(-23)Jmol^(-1)K^(-1)` T=300 K `K=(3)/(2)K_(B)T` `therefore K=(3)/(2)xx1.38xx10^(-23)xx300` `therefore K=621xx10^(-23)J` `therefore` de-Broglie wavelength, `lambda=(h)/(sqrt(2mK))` `therefore lambda=(6.63xx10^(-34))/(sqrt(2xx1.67xx10^(-27)xx621xx10^(-23)))` `therefore lambda=(6.63xx10^(-34))/(45.54xx10^(-25))` `therefore lambda=0.14558xx10^(-9)m` `therefore lambda~~0.146` nm |
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