Question

The de-Broglie wavelength of a particle accelerated with a p.d of 150 V is ${10}^{-10}m$. If it is accelerated with a p.d of 600 V, its wavelength will be

• A. $0.25\stackrel{\circ }{A}$
• B. $0.5\stackrel{\circ }{A}$
• C. $1.5\stackrel{\circ }{A}$
• D. $2\stackrel{\circ }{A}$

Answer 1

The correct option is B $0.5\stackrel{\circ }{A}$

$\frac{1}{2}m{v}^2=qV\Rightarrow mv=\sqrt{28Vm}\lambda =\frac{h}{p}=\frac{h}{mv}=\frac{h}{\sqrt{2mqV}}\Rightarrow \lambda \alpha \frac{1}{\sqrt{V}}\Rightarrow \frac{{\lambda }_1}{{\lambda }_2}=\sqrt{\frac{V_2}{V_1}}\Rightarrow \frac{{10}^{-10}}{{\lambda }_2}=\sqrt{\frac{600}{150}}=2\Rightarrow {\lambda }_2=0.5A^{\circ }.$