A deuteron and an alpha particle are accelerated from rest through...

The de-Broglie wavelength of a particle accelerated through a potential difference $V$ is given by $\lambda = \dfrac{h}{\sqrt{2mqV}}$, where $m$ is the mass and $q$ is the charge of the particle.

Let the mass of a proton be $m$ and its charge be $e$.

For a deuteron:
Mass $m_1 = 2m$
Charge $q_1 = e$
Wavelength $\lambda_1 = \dfrac{h}{\sqrt{2(2m)(e)V_1}} = \dfrac{h}{\sqrt{4meV_1}}$

For an alpha particle:
Mass $m_2 = 4m$
Charge $q_2 = 2e$
Wavelength $\lambda_2 = \dfrac{h}{\sqrt{2(4m)(2e)V_2}} = \dfrac{h}{\sqrt{16meV_2}}$

Given that the de-Broglie wavelengths are equal ($\lambda_1 = \lambda_2$):

$\dfrac{h}{\sqrt{4meV_1}} = \dfrac{h}{\sqrt{16meV_2}}$

Squaring both sides, we get:

$4meV_1 = 16meV_2$

$\dfrac{V_1}{V_2} = \dfrac{16}{4} = \dfrac{4}{1}$

The ratio $\dfrac{V_1}{V_2}$ is $4:1$.

Answer: $4:1$