Question

The diameter of the nucleus is about $10\mathrm{fm}$. What is the kinetic energy, in MeV, of a proton with a de Broglie wavelength of$10\mathrm{fm}$ ?

Short answer

$E_k=1.3·{10}^{-12}\mathrm{J}=8.2\mathrm{MeV}$

Step-by-step solution

Given Information

The diameter of the nucleus is about $10\mathrm{fm}$. proton with a de Broglie wavelength of $10\mathrm{fm}$

Calculation

We can start the solution by determining de Broglie wavelength

$$\begin{gathered} \lambda =\frac{h}{mv} \\ v=\frac{h}{m\lambda } \\ {E}_k=\frac{m{v}^2}{2} \\ {E}_k=\frac{m{h}^2}{2m^2{\lambda }^2} \\ {E}_k=\frac{h^2}{2m{\lambda }^2} \\ {E}_k=\frac{\left({\left(6.63·{10}^{-34}\right)}^2\right.}{2·1.67·{10}^{-27·{\left(10·{10}^{-15}\right)}^2}} \\ {E}_k=1.3·{10}^{-12}\mathrm{J}=8.2\mathrm{MeV} \end{gathered}$$