In a cyclotron, if a deuteron can gain an energy of 40 MeV, then a proton can gain an energy of

Options

A. 40 MeV
B. 80 MeV
C. 20 MeV
D. 60 MeV

Answer

B

Explanation

As we know,
`F=qVB=(mv^(2))/( r )` = contripetal force maximum energy, `E=(1)/(2)(B^(2)q^(2)r^(2))/(m)`
Hence ratio of energy of deuteron and proton is `(E_(d))/(E_(p))=((q_(d))/(q_(p)))^(2)((m_(p))/(m_(d)))`
`(40)/(E_(p))=((q)/(q))^(2)(m)/(2m)`
`E_(p)=80" Mev"`