An alpha particle is accelerated through a potential difference of 100 V. Calculate: (i) The speed acquired by the alpha particle, and (ii) The…

$\displaystyle \mathrm{V}=\sqrt{\frac{2 q V}{m}}$

$$\begin{aligned} & =\sqrt{\frac{$\displaystyle 2$ \times $\displaystyle 2$ \times $\displaystyle 1.6$ \times $\displaystyle 10$^{-$\displaystyle 19$} \times $\displaystyle 100$}{$\displaystyle 6.4$ \times $\displaystyle 10$^{-$\displaystyle 27$}}} & \quad=$\displaystyle 1$ \times $\displaystyle 10$^{$\displaystyle 5$} \mathrm{~m} / \mathrm{s} \end{aligned}

$$ii) $\displaystyle \lambda=\frac{h}{p}=\frac{h}{m v}$

$$\begin{aligned} & =\frac{$\displaystyle 6.63$ \times $\displaystyle 10$^{-$\displaystyle 34$}}{$\displaystyle 6.4$ \times $\displaystyle 10$^{-$\displaystyle 27$} \times $\displaystyle 10$^{$\displaystyle 5$}} & \lambda=$\displaystyle 10$^{-$\displaystyle 12$} \mathrm{~m} \end{aligned}