Answer
$5.236\;fm$
Work Step by Step
The de Broglie wavelength $(\lambda)$ is express by the formula,
$\lambda=\frac{h}{p}$, ......................$(1)$
where $h$ is the planck's constant and $p$ is the momentum of the moving particle.
Again, the relation between $p$ and kinetic energy $K$ is
$K=\frac{p^2}{2m}$ , where $m$ is the mass of the particle.
Thus, $p=\sqrt {2mK}$ ......................$(2)$
Substituting eq. $2$ in eq. $1$, we get
$\lambda=\frac{h}{\sqrt {2mK}}$ ......................$(3)$
Given $K=7.5\;MeV$
The mass alpha particle is around four times greater than the mass of proton. Therefore, the mass alpha particle is $m_{\alpha}=4\times 1.67\times 10^{-27}\;kg$
Thus, the de Broglie wavelength of the alpha particle is given by
$\lambda_{\alpha}=\frac{h}{\sqrt {2m_{\alpha}K}}$
or, $\lambda_{\alpha}=\frac{6.63\times 10^{-34}}{\sqrt {2\times 4\times 1.67\times 10^{-27}\times 7.5\times 10^{6}\times 1.6\times 10^{-19}}}$
or, $\lambda_{\alpha}\approx5.236\times 10^{-15}\;m$
or, $\lambda_{\alpha}=5.236\;fm$
