Physics - Modern Physics Question with Solution | TestHub

Since, we know de-Broglie wavelength,

$\lambda =\frac{h}{p}$

And$p=\sqrt{2m_{p}KE}$

$=\begin{array}{cc}\sqrt{2m_p\frac{3k}{2}\cdot T}& \left[\because KE=\frac{3k}{2}\cdot T\right]\end{array}$

$=\sqrt{3m_{p}kT}$

So,$\lambda =\frac{h}{\sqrt{3m_{p}kT}}$

Putting the given values, we get

$\therefore \lambda =\frac{6.63\times {10}^{-34}}{\sqrt{3\times 1.67\times {10}^{-27}\times 1.38\times {10}^{-23}\times T}}$

$=\frac{2.52}{\sqrt{T}}\times {10}^{-9} \mathrm{m}$

or$\lambda$$=\frac{25.2}{\sqrt{T}}\times {10}^{-10} \mathrm{m}$

$\therefore \lambda =$$\frac{25.2}{\sqrt{T}} \stackrel{\circ }{\mathrm{A}}$