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 B where they move in circular paths perpendicular to B. Determine the radius of the paths for the deuteron and alpha particle in terms of that for the proton.

Answer 1

Answer with Explanation:

We are given that

Mass of deuteron=$2m_p$

Charge, Q=e

Mass of alpha particle=$4m_p$

Charge,q=2e

Magnetic field=B

Mass of proton=$m_p$

Let radius of path of proton=r

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

Using the formula

Velocity of proton=$v=\sqrt{\frac{2qV}{m}}$

Centripetal force =Magnetic force

$\frac{mv^2}{r}=qvB$

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

Radius of proton,$r=\frac{m_p\times\sqrt{\frac{2eV}{m_p}}}{eB}=\frac{\sqrt{2V}}{B}\sqrt{\frac{m_p}{e}}$

Radius of deuteron,$R=\frac{\sqrt{2V}}{B}\times \sqrt{\frac{2m_p}{e}}$

$\frac{R}{r}=\sqrt{2}$

$R=\sqrt{2}r$

Radius of alpha particle,$R'=\frac{\sqrt{2V}}{B}\times\sqrt{\frac{4m_p}{2e}}$

$\frac{R'}{r}=\sqrt 2$

$R'=\sqrt{2} r$