A proton carrying 1 MeV kinetic energy is moving in a circular path of radius R in uniform magnetic field. What should be the energy of an `alpha`-particle to describe a circle of same radius in the same field?

Options

A. 2 MeV
B. 1 MeV
C. 0.5 MeV
D. 4 MeV

Answer

B

Explanation

Kinetic energy of a charged particle
`K = 1/2 mv^2 or v = sqrt((2K)/(m))`
Radius of the circular path of charged particle in unifrom magnetic field is given by
`R = (mv)/(Bq) = (m)/(Bq) sqrt((2K)/(m)) = sqrt((2mK)/(Bq))`
Mass of a proton `m_p = m`
Charge of a proton, `m_alpha = 4m `
Charge of a proton, `q_p = e`
Charge of an `alpha - particle q_alpha = 2e`
`therefore R_P = (sqrt(2m_p K_p))/(Bqalpha) = sqrt(2mK_p)/(Be)`
and `R_(alpha) = sqrt(2m_alpha K_alpha)/(Bq_alpha)=sqrt(2(4m)K_alpha)/(B(2e))= sqrt(2mK_alpha)/(Be)`
`therefore (R_p)/(R_alpha) = sqrt((K_p)/(K_alpha))`
As `R_p = R_alpha` (Given)
`therefore K_alpha = K_p = 1 MeV`