Question

A deuteron and an alpha particle are accelerated with the same accelerating potential greater value of de-Broglie wavelength, associated it ?

Answer 1

de-Broglie wavelength of a particle depends upon its mass and charge for same accelerating potential, such that

\[\lambda \propto \frac{1}{\sqrt{(\text { mass })(\text { charge })}}\]

Mass and charge of a deuteron are 2m_p and e respectively,
and, mass and charge of an alpha particle are 4m_p and 2e respectively.
where, m_p is the mass of a proton and e is the charge of an electron

\[\frac{\lambda_D}{\lambda_\alpha} = \sqrt{\frac{m_\alpha q_\alpha}{m_D q_D}} = \sqrt{\frac{(4 m_p )(2e)}{(2 m_p )(e)}} = \frac{2}{1}\]

Thus, de-broglie wavelength associated with deutron is twice of the de-broglie wavelength of alpha particle.