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a. For what kinetic energyof a neutron will the associated de Broglie wavelength be 1.40xx10^(-10)m? b. Also find the de Broglie wavelengthof a neutron, the thermal equilibrium with matter, having an average kinetic energyof (3/2) k T at 300 K. |
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Answer» Solution :`lambda=1.4xx10^(-10)m` a. `K_("max")=(1)/(2)mv^(2)=(p^(2))/(2m)=(h^(2))/(2m lambda^(2))=((6.6xx10^(-34))^(2))/(2xx1.67xx10^(-27)XX(1.4xx10^(-10))^(2))=6.65x10^(-21)J` b.`lambda=(h)/(sqrt(3mKT))=(6.6xx10^(-34))/(sqrt(3xx1.6xx10^(-27)xx1.38xx10^(-23)xx300))=(6.6xx10^(-34))/(sqrt(1987.2)xx10^(25))=(6.6)/(44.5)xx10^(-9)` `=0.148xx10^(-9)m=0.148nm` |
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