A proton and an alpha particle are accelerated through the same potential. Which one of the two has (i) greater value of de−Broglie wavelength associated with it and (ii) less kinetic energy. Give reasons to justify your answer.

#### Solution

(a) 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 proton are m_{p} and e respectively,

and, mass and charge of an alpha particle are 4m_{p} and 2e respectively.

where, e is the charge of an electron

\[\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{m_\alpha q_\alpha}{m_p q_p}} = \sqrt{\frac{(4 m_p )(2e)}{( m_p )(e)}} = 2\sqrt{2}\]

Thus, de-broglie wavelength associated with proton is

Charge of an alpha particle is more as compared to a proton. So, it will have a greater value of K.E. Hence, proton will have lesser kinetic energy.