1.

The de-Broglie wavelength of a neutron at 27^(@)C is lamda, what will be the corresponding wavelength at 927^(@)C ?

Answer»

`lamda/2`
`lamda/3`
`lamda/4`
`lamda/9`

Solution :SINCE `lamda=(H)/(sqrt(2mK)) and K=(3)/(2)k_(B)T`
`implies lamda=(h)/(sqrt(3mk_(B)T)) or lamda PROP (1)/(sqrtT)`
`therefore lamda_(2)=lamda_(1)sqrt((T_(1))/(T_(2)))=sqrt((273+27)/(273+927))lamda=(lamda)/(2)`.


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