The de Broglie wavelength of a neutron is 144 pm. What is its velocity ?

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Answer 1 · 2016-11-09T00:16:43

$sf(v=2.77xx10^(3)color(white)(x)"m/s")$

$sf(lambda=h/(mv))$

h is the Planck constant = $sf(6.63xx10^(-34)color(white)(x)Js)$

$sf(lambda)$ is the wavelength = 144 pm = $sf(1.44xx10^(-10)color(white)(x)m)$

$sf(m=1.67xx10^(-27)color(white)(x)kg)$

$:.$ $sf(v=h/(mlambda)$

$sf(v=(6.63xx10^(-34))/(1.67xx10^(-27)xx1.44xx10^(-10))color(white)(x)"m/s")$

$sf(v=2.77xx10^(3)color(white)(x)"m/s")$