Explanation:
Charge on proton, q₁ = e
Charge on alpha particles, q₂ = 2e
The magnetic force is given by :
[tex]F=qvB\ sin\theta[/tex]
Here, [tex]\theta=90=sin(90) = 1[/tex]
For proton, [tex]F_p=ev_pB[/tex]..........(1)
For alpha particle, [tex]F_a=2ev_aB[/tex]..........(2)
Since, a proton (charge e), traveling perpendicular to a magnetic field, experiences the same force as an alpha particle. So,
[tex]ev_pB=2ev_aB[/tex]
[tex]\dfrac{v_p}{v_a}=\dfrac{2}{1}[/tex]
So, the ratio of the speed of proton to the alpha particle is 2 : 1 .Hence, this is the required solution.
