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Calculate the de Broglie wavelength of a neutron of energy 0.022 eVm = 1.67 xx 10^(-27) kg. |
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Answer» SOLUTION :GIVEN `MV^2=0.022eV=0.022xx1.6xx10^-19J` `m^2v^2=2mxx0.022xx1.6xx10^-19` `mv=SQRT(2xxmxx0.022xx1.6xx10^-19` `lambda=lambda/(mv)=(6.62xx10^-34)/sqrt(2xx1.67xx10^-27xx0.022xx1.6xx10^-19)=1.93xx10^-10m` |
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