The nuclear radius of ${ }_{13}^{27}\mathrm{Al}$ is 3·6 fermi. Find the nuclear radius of ${ }_{29}^{64}\mathrm{Cu}$
OR
A proton and an electron have equal speeds. Find the ratio of de Broglie wavelength associated with them.

We know that, $\mathrm{r}={\mathrm{r}}_0{\mathrm{A}}^{1/3}$
So, ${\mathrm{r}}_{\mathrm{Cu}}/{\mathrm{r}}_{\mathrm{Al}}={\mathrm{A}}_{\mathrm{Cu}}^{1/3}/{\mathrm{A}}_{\mathrm{Al}}^{1/3}={64}^{1/3}/{27}^{1/3}=4/3$
or, ${\mathrm{r}}_{\mathrm{Cu}}=4/3\times 3.6=4.8 \mathrm{Fermi}$
OR 
wavelengths associated with them.
Taking the ratio of the wavelengths,
${\mathrm{\lambda }}_1:{\mathrm{\lambda }}_2=\mathrm{h}/{\mathrm{m}}_1{\mathrm{V}}_1:\mathrm{h}/{\mathrm{m}}_2{\mathrm{V}}_2$
Canceling the common terms,
$\begin{array}{l}{\mathrm{\lambda }}_1:{\mathrm{\lambda }}_2={\mathrm{m}}_2:{\mathrm{m}}_1\\ {\mathrm{\lambda }}_1:{\mathrm{\lambda }}_2=\left(1.67262192\times {10}^{-27}\right)\mathrm{kg}/\left(9.1093837\times {10}^{-31}\right)\mathrm{kg}\end{array}$
${\mathrm{\lambda }}_1:{\mathrm{\lambda }}_2=1836:1$
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