Estimate the wavelength of electrons that have been accelerated from rest through a potential difference of $40 \mathrm{kV}$.
 The expression $p^2 / 2 m_{\mathrm{e}}=e \Delta \phi$ solves to $p=\left(2 m_{\mathrm{e}} e \Delta \phi\right)^{1 / 2}$; then, from the de Broglie relation $\lambda=h / p$,
$$
\lambda=\frac{h}{\left(2 m_{\mathrm{e}} e \Delta \phi\right)^{1 / 2}}
$$
Substitution of the data and the fundamental constants (from inside the front cover) gives
$$
\begin{aligned}
\lambda & =\frac{6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}}{\left\{2 \times\left(9.109 \times 10^{-31} \mathrm{~kg}\right) \times\left(1.602 \times 10^{-19} \mathrm{C}\right) \times\left(4.0 \times 10^4 \mathrm{~V}\right)\right\}^{1 / 2}} \\
& =6.1 \times 10^{-12} \mathrm{~m}
\end{aligned}
$$
