Question

An $\alpha$ particle is accelerated through a potential difference of $Vvolts$. The de Broglie's wavelength associated with it is:

• A. $\sqrt{\frac{150}{V}}\stackrel{\circ }{A}$
• B. $\frac{0.286}{\sqrt{V}}\stackrel{\circ }{A}$
• C. $\frac{0.101}{\sqrt{V}}\stackrel{\circ }{A}$
• D. $\frac{0.983}{\sqrt{V}}\stackrel{\circ }{A}$

Answer 1

The correct option is C $\frac{0.101}{\sqrt{V}}\stackrel{\circ }{A}$

Given:
Applied potential difference = $Vvolts$

We know that
$\text{Mass of}\alpha \text{particle}\left(m\right)=6.64\times {10}^{-27}kg$
$\text{Charge on}\alpha \text{particle}=2e(e=1.6\times {10}^{-19}coulomb)$

Hence final kinetic energy of the $\alpha$ particle $\left(KE\right)=V\times 2e$
de Broglie wavelength$\left(\lambda \right)$ in terms of kinetic energy (KE) can be given as:
$\lambda =\frac{h}{\sqrt{2KEm}}$
Since
$KE=V\times 2e⟹\lambda =\frac{h}{\sqrt{2\times V\times 2e\times m}}=\frac{6.63\times {10}^{-34}}{\sqrt{4\times V\times 1.6\times {10}^{-19}\times 6.64\times {10}^{-27}}}=\frac{1.01\times {10}^{-11}}{\sqrt{V}}m=\frac{0.101}{\sqrt{V}}\stackrel{\circ }{A}$
So c is the answer.