Question

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

• A. $\frac{25.2}{\sqrt{T}}\stackrel{{}^{\circ }}{A}$
• B. $\frac{0.308}{\sqrt{T}}\stackrel{{}^{\circ }}{A}$
• C. $\frac{0.025}{\sqrt{T}}\stackrel{{}^{\circ }}{A}$
• D. $\frac{0.25}{\sqrt{T}}\stackrel{{}^{\circ }}{A}$

Answer 1

Answer: (A)

Since, we know de-Broglie wavelength,
$\lambda =\frac{h}{p}$
And $p=\sqrt{2m_{p}KE}$
$=\sqrt{2m_p\frac{3k}{2}\cdot T}\left[\because KE=\frac{3k}{2}\cdot T\right]$
$=\sqrt{3m_{p}kT}$
So, $\lambda =\frac{h}{\sqrt{3m_{p}kT}}$
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 $\lambda$ $=\frac{25.2}{\sqrt{T}}\times 10^{-10}m$
$\therefore \lambda =$ $\frac{25.2}{\sqrt{T}}\stackrel{{}^{\circ }}{A}$