A proton and an `alpha`-particle are accelerated, using the same potential difference. How are the de-Broglie wavelengths `lambda_(p) and lambda_(a)` related to each other?

Explanation

For a particle of mass m , charge q , accelerated through a potential V, the de Broglie wavelength is given by
`lamda=h/(sqrt(2mqV))`
For fixed V ,
`lamdaprop1/(sqrt(mq))`
Ratio of the Broglie wavelength for proton and alpha particle is
`lamda_p/lamda_(alpha)=(sqrt(m_alphaq_alpha))/(sqrt(m_pq_p))`
We know that `m_alpha=4m+p,q_p=e and q_(alpha)=2e`
`:.lamda_p/lamda_alpha=sqrt(4m_pxx2e)/sqrt(m_pxxe)=sqrt8/1`