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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.

Respuesta :

lublana

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]