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.
Question
Answer with Explanation:
We are given that
Mass of deuteron=[tex]2m_p[/tex]
Charge, Q=e
Mass of alpha particle=[tex]4m_p[/tex]
Charge,q=2e
Magnetic field=B
Mass of proton=[tex]m_p[/tex]
Let radius of path of proton=r
[tex]v=\sqrt{\frac{2qV}{m}}[/tex]
Using the formula
Velocity of proton=[tex]v=\sqrt{\frac{2qV}{m}}[/tex]
Centripetal force =Magnetic force
[tex]\frac{mv^2}{r}=qvB[/tex]
[tex]r=\frac{mv}{qB}[/tex]
Radius of proton,[tex]r=\frac{m_p\times\sqrt{\frac{2eV}{m_p}}}{eB}=\frac{\sqrt{2V}}{B}\sqrt{\frac{m_p}{e}}[/tex]
Radius of deuteron,[tex]R=\frac{\sqrt{2V}}{B}\times \sqrt{\frac{2m_p}{e}}[/tex]
[tex]\frac{R}{r}=\sqrt{2}[/tex]
[tex]R=\sqrt{2}r[/tex]
Radius of alpha particle,[tex]R’=\frac{\sqrt{2V}}{B}\times\sqrt{\frac{4m_p}{2e}}[/tex]
[tex]\frac{R’}{r}=\sqrt 2[/tex]
[tex]R’=\sqrt{2} r[/tex]