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Medical
1 year, 1 month ago

# Solve it, A proton and an alpha particle both enter a region of uniform magnetic field B, moving at right angles to the field B. If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is 1 MeV, the

A proton and an alpha particle both enter a region of uniform magnetic field B, moving at right angles to the field B. If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is 1 MeV, the energy acquired by the alpha particle will be:

• Option 1)

0.5 MeV

• Option 2)

1.5 MeV

• Option 3)

1 MeV

• Option 4)

4 MeV

899 Views
V Vakul Arora
Answered 1 year, 1 month ago

$R=\frac {mv}{qB} = \frac {\sqrt{2mk}}{qB}$$K_{p}=\frac {q^{2}B^{2}R_{p}^{2}}{2mp},\: K\infty = \frac{q_{\infty}^{2}B^{2}R_{\infty}^{2}}{2m\infty}$

$K=\frac {q^{2}B^{2}R^{2}}{2m}$

$\therefore \frac{K\infty}{K_{p}}=\left ( \frac{q\infty}{q_{Io}} \right )^{2} \left ( \frac{mp}{m\infty} \right )\left ( \frac{Rp}{R\infty} \right )^{2}$

=>$K_{\infty}=K_{p} \left ( \frac{q_{\infty}}{q_{p}} \right )^{2}\left ( \frac{mp}{m\infty} \right )\left ( \frac{R_{\infty}}{R_{p}} \right )^{2}$

Kp=1MeV$(2)^{2} (\frac{1}{4}) (1)^{2}=1mev$

Option 1)

0.5 MeV

Option 2)

1.5 MeV

Option 3)

1 MeV