# The de-Broglie wavelength of neutron in thermal equilibrium at tempreture T is Option 1) $\frac{30}{\sqrt{T}}A^{\circ}$ Option 2) $\frac{3.08}{\sqrt{T}}A^{\circ}$ Option 3) $\frac{0.308}{\sqrt{T}}A^{\circ}$ Option 4) $\frac{0.0308}{\sqrt{T}}A^{\circ}$

P Plabita

As we discussed in concept

De - Broglie wavelength with charged particle -

$\lambda = \frac{h}{\sqrt{2mE}}= \frac{h}{\sqrt{2mE}}$

$\lambda = \frac{h}{\sqrt{2mqv}}$

- wherein

$E\rightarrow kinetic\: energy\: o\! f particle$

$q\rightarrow charged \: particle$

$\lambda\:=\:\frac{h}{p}=\frac{h}{\sqrt{2m}E}=\frac{h}{\sqrt{2m}.(\frac{3}{2}KT)}=\frac{h}{\sqrt{3mK}T}=\frac{30}{\sqrt{T}}A^{\circ}$

Option 1)

$\frac{30}{\sqrt{T}}A^{\circ}$

This option is correct.

Option 2)

$\frac{3.08}{\sqrt{T}}A^{\circ}$

This option is incorrect.

Option 3)

$\frac{0.308}{\sqrt{T}}A^{\circ}$

This option is incorrect.

Option 4)

$\frac{0.0308}{\sqrt{T}}A^{\circ}$

This option is incorrect.

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