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The de Broglie wavelength of a thermal n...

The de Broglie wavelength of a thermal neutron at `927^circC` is `lamda`. Its wavelength at `327^circC` will be

A

`(lamda)/(2)`

B

`(lamda)/(sqrt2)`

C

`lamdasqrt2`

D

`2lamda`

Text Solution

Verified by Experts

The correct Answer is:
C
`lamdaprop(1)/(V)` and `VpropsqrtT`
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