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The de - Broglie wavelength of a neutron...

The de - Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature `T ("kelvin")` and mass `m`, is

A

`h/(sqrt(mkT))`

B

`h/(3sqrt(mkT))`

C

`(2h)/(sqrt(3mkT))`

D

`(2h)/(sqrt(mkT))`

Text Solution

Verified by Experts

The correct Answer is:
B
The de Broglie wavelength of the
`lamda=h/(mv)=h/(sqrt(2m(K.E)))`
where K.E. is the kinetic energy.
Thus, `lamda=h/(sqrt(2m(3/2kT)))`
`lamda=h/(sqrt(3mkT))`
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