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The de-Brogile wavelength of a neutron a...

The de-Brogile wavelength of a neutron at `927^(@)`C is `lamda`. What will be its wavelength at `27^(@)`C?

A

`lamda//2`

B

`lamda`

C

`2lamda`

D

`4lamda`

Text Solution

Verified by Experts

The correct Answer is:
3
`lamda=h/(mv)" "vpropsqrtT`
`lamda=h/(mxxsqrt(1200))`
`lamda_(1)=h/(mxxsqrt(300))`
`eq. 2/1`
`lamda_(1)/lamda_(2)=sqrt(1200/300)`
`lamda_(1)=2lamda`
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