A proton and deutron are accelerated through the same accelerating potential. Which one of the two has (a) greater value of de-broglie wavelength associated with it, and (b) less momentum? Give reasons to justify your answer.

Explanation

`lamda=h/(sqrt(2mqV))`
For same V :
`lamda_p//lamda_d=sqrt((m_pq_p)/(m_pq_p))=sqrt((m_pq_p)/(2m_pq_p))=sqrt(1/2)`
`:.m_d=2m_p`
`q_d=q_p`
`lamda_p=sqrt2lamda_alpha`
Thus proton has higher de Broglie wavelength than deuteron.
(b) `p=h/lamda`
Higher the wavelength , less is the momentum.
So , proton having larger wavelength will have less momentum.