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For what kinetic energy of a neutron will the associated de-Broglie wavelength be `1.40 xx 10^(-10)m` ? Mass of neutron = `1.675 xx 10^(-27) kg, h = 6.63 xx 10^(-34) Js`. |
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Answer» Givn, for neutron, `lambda = 1.40xx10^(-10)m` and `m = 1.675xx10^(-27)kg` `KE = (P^(2))/(2m)=(h^(2))/(2m lambda^(2))= ((6.63xx10^(-34))^(2))/(2xx(1.40xx10^(-10))^(2)xx1.675xx10^(-27))therefore KE = 6.686xx10^(-21)J` |
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