Question

Show that the maximum wavelength associated with an electron accelerated by potential V is given by $\lambda = \frac{12.27}{\sqrt{V}} A^\circ$.

Answer 1

The maximum wavelength associated with an electron accelerated by potential $V$ is given by $\lambda = \frac{12.27}{\sqrt{V}} \text{ Å}$.

Answer 2

An electron accelerated by potential $V$ gains kinetic energy $KE = eV$. Its momentum is $p = \sqrt{2mKE} = \sqrt{2meV}$. By de Broglie's hypothesis, $\lambda = h/p$. Substituting $p$, we get $\lambda = h/\sqrt{2meV}$. The constant $\frac{h}{\sqrt{2me}}$ evaluates to approximately $1.226 \times 10^{-9}$ m, which is $12.26 \text{ Å}$. Thus, $\lambda = \frac{12.26}{\sqrt{V}} \text{ Å}$, commonly rounded to $12.27 \text{ Å}$.