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

Options

A. `lamda/2`
B. `lamda`
C. `2lamda`
D. `4lamda`

Answer

C

Explanation

de-broglie wavelength of a material particle at temperature T is given by
`lamda = (h)/(sqrt(2mkT )) `, where k is boltzmann.s constant
`rArr lamda prop (1)/(sqrtT)`
` therefore lamda_2/lamda_1 = sqrt(T_1/T_2)`
`lamda_2/lamda_1 = sqrt(1200/300) = 2`
` therefore lamda_2 = 2lamda_1 = 2 lamda`