Question

The de-Broglie wavelength of neutrons in thermal equilibrium at temperature $T$ is

• A. $\frac{25 .2}{\sqrt{T}}\overset{^\circ }{A}$
• B. $\frac{0 . 308}{\sqrt{T}}\overset{^\circ }{A}$
• C. $\frac{0 . 025}{\sqrt{T}}\overset{^\circ }{A}$
• D. $\frac{0 . 25}{\sqrt{T}}\overset{^\circ }{A}$

Answer 1

Answer: (A)

Since, we know de-Broglie wavelength,
$\lambda =\frac{h}{p}$
And $p=\sqrt{2 m_{p} K E}$
$=\sqrt{2 m_{p} \frac{3 k}{2} \cdot T} & \left[\because K E = \frac{3 k}{2} \cdot T\right]$
$=\sqrt{3 m_{p} k T}$
So, $\lambda =\frac{h}{\sqrt{3 m_{p} k T}}$
Putting the given values, we get
$\therefore \, \, \lambda =\frac{6.63 \times 10^{- 34}}{\sqrt{3 \times 1.67 \times 10^{- 27} \times 1.38 \times 10^{- 23} \times T}}$
$=\frac{2.52}{\sqrt{T}}\times 10^{- 9}m$
or $\lambda $ $=\frac{25.2}{\sqrt{T}}\times 10^{- 10}m$
$\therefore \, \, \lambda =$ $\frac{25.2}{\sqrt{T}}\overset{^\circ }{A}$