Question

A proton (charge +e, mass mp), a deuteron (charge +e, mass 2mp), and an alpha particle (charge +2e, mass 4mp) are accelerated from rest through a common potential difference ?V. Each of the particles enters a uniform magnetic field B(-->) , with its velocity in a direction perpendicular to B(-->). The proton moves in a circular path of radius rp.
(a) In terms of rp, determine the radius rd of the circular orbit for the deuteron.
rd = _________



(b) In terms of rp, determine the radius r(aplha) for the alpha particle.

r(alpha)________________

Answer 1

Answer:

(a) $r_{d}=\sqrt{2}\times r_{P}$

(b) $r_{\alpha}=\sqrt{2}\times r_{P}$

Explanation:

charge on proton, q = e

mass pf proton, m = mp  

charge on deuteron, q' = e

mass pf deuteron, m' = 2mp  

charge on alpha particle, q'' = 2e

mass pf alpha particle, m'' = 4mp

Potential difference = V  

Magnetic field = B

The kinetic energy is given by eV.

So, 1/2 mv^2 = e V

where, v be the velocity of the particle.

$v=\sqrt{\frac{2eV}{m}}$

The formula for the radius of circular path is given by

$r = \frac{mv}{Bq}$

By substituting the value of v

$r = \frac{\sqrt{2eVm}}{Bq}$

$r\alpha \frac{\sqrt{m}}{q}$

The radius of path of proton is given by

$r_{p}\alpha \frac{\sqrt{m_{p}}}{e}$.... (1)

(a) radius of deuteron

$r_{d}\alpha \frac{\sqrt{2m_{p}}}{e}$

By comparing equation (1), we get

$r_{d}=\sqrt{2}\times r_{P}$

(b) radius of alpha particle

$r_{\alpha}\alpha \frac{\sqrt{4m_{p}}}{2e}$

By comparing equation (1), we get

$r_{\alpha}=\sqrt{2}\times r_{P}$