Question

A stream of protons and deuterons in a vacuum chamber enters a uniform magnetic field. Both protons and deuterons have been subjected to same accelerating potential, hence the kinetic energies of the particles are the same. If the ion-stream is perpendicular to the magnetic field and the protons move in a circular path of radius 15 cm, find the radius of the path traversed by the deuterons. Given that mass of deuteron is twice that of a proton.

Answer 1

The radius `r` of the path is given by
`(mv^2)/r=qvB or r=(mv)/(qB)`
For proton, `r_p=(m_pv_p)/(q_pB)`
For deuteron, `r_d=(m_dv_d)/(q_dB)`
`:. (r_d)/(r_p)=(m_dv_d)/(m_pv_p)xx(q_pB)/(q_dB)`….(i)
Given that `1/2m_pv_p^2=1/2m_dv_d^2` also `q_b=q_d`
Now `(v_d)/(v_p)=((m_p)/(m_d))^(1//2)`.....(ii)
Substituting the value of `v_d//v_p` from Eq. (i) in Eq.(i) we get
`(r_d)/(r_p)=(m_d)/(m_p) ((m_p)/(m_d))^(1//2)=((m_d)/(m_p))^(1//2)=sqrt2 (because m_d=2m_p)`
or `r_d=sqrt2.r_p=1.414xx0.15=0.212m`