#### Question

An a-particle and a proton of the same kinetic energy are in turn allowed to pass through a magnetic field `vec"B"`, acting normal to the direction of motion of the particles. Calculate the ratio of radii of the circular paths described by them.

#### Solution

Mass of alpha particle = `m_a`

Charge on alpha particle = `q_α`

Velocity of alpha particle = `V_α`

Mass of a proton = m_{p }

Charge on a proton = q_{p }

Velocity of proton = v_{p}

Since the KE of alpha particle and proton are equal.

∴ `1/2m_αν_α^2 = 1/2m_p.ν_p^2`

⇒ `(m_αν_α)^2/m_α = (m_p.ν_p)^2/m_p ⇒ (m_αν_α)/(m_p.ν_p) = (m_α/m_p)^(1/2)`

Radius of the circle = R = `(mν)/(qB)`

Therefore, ratio of radii of alpha particle to proton

= `R_α/R_p = (m_α.ν_α)/(q_αB).(q_pB)/(m_p.ν_p) = (m_α/m_p)^(1/2).q_p/q_α`

Using `m_alpha/m_p = 4` and `q_p/q_alpha = 1/2`

∴ `R_alpha/R_p = (4)^(1/2)(1/2) = 1`