Question

A proton (mass mp), a deuteron (m=2mp,Q=e), and an alpha particle (m=4mp,Q=2e) are accelerated by the same potential difference V and then enter a uniform magnetic field B⃗ , where they move in circular paths perpendicular to B⃗ .

Determine the radius of the path for the deuteron in terms of that for the proton.

Determine the radius of the path for the alpha particle in terms of that for the proton.

Express your answers in terms of rp.

Answer 1

At equilibrium, Lorentz force is balanced by the centripetal force given by

  

where q= charge on particle

v= velocity of the particle

B= Magnetic Field

m= mass of the particle

r= radius of the circle followed by the particle in External Magnetic Field

................................(1)

Now Using Energy Conservation Principle,

Kinetic Energy= Potential Energy

where V= Potential applied to the particle,

  

then using above equation in equation 1

  

Since the potential applied(V) and Magnetic Field(B) is same for all the particle hence ratio of the radius of the particle to the radius of proton will be

a). For Deutron, m'= 2m_p and q'= q_p, then ratio of the radius will be

  

  (ANS)

b). For Alpha particle, m'= 4m_p and q'= 2q_p, then ratio of the radius will be

  

  (ANS)