A proton and an `alpha` particle are accelerated through the same potential difference V. The ratio of their de Broglie wavelengths is

Options

A. `sqrt2`
B. `2sqrt2`
C. `sqrt3`
D. `2sqrt3`

Answer

B

Explanation

de Broglie wavelength of a particle of mass m and charge q accelerated through a potential difference V is
`lamdah/(sqrt(2mqV))`
Since potential difference V is the same for the the particles,
`lamdaproph/(sqrt(mq))`
Thus, `lamda_p/lamda_alpha=sqrt((m_alphaq_alpha)/(m_pq_p))=sqrt(((4m_p)/m_p)((2e)/e))=sqrt8/1=2sqrt2`