Cyclotrons are used to accelerate ions like deuterons (\( d \)) and \( \alpha \) particles. Keeping the magnetic field the same for both, \( d \) and \( \alpha \) are extracted with energies 10 MeV and 20 MeV with extraction radii \( r_d \) and \( r_\alpha \), respectively. Taking the masses \( M_d = 2000 \, {MeV}/c^2 \) and \( M_\alpha = 4000 \, {MeV}/c^2 \), the value of \( \frac{r_\alpha}{r_d} \) (in integer) is:
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Solution and Explanation
The energy of a charged particle in a cyclotron is related to its momentum, and the radius of the circular trajectory is given by:
\[ r = \frac{mv}{qB} \] where \( m \) is the mass of the particle, \( v \) is its velocity, \( q \) is the charge of the particle, and \( B \) is the magnetic field.
For the cyclotron, the kinetic energy \( E \) of the particle is related to its momentum \( p \) by:
\[ E = \frac{p^2}{2m} \]
Thus, the momentum of the particle is:
\[ p = \sqrt{2mE} \]
Now, using the relation for the radius:
\[ r = \frac{p}{qB} = \frac{\sqrt{2mE}}{qB} \]
For deuterons (\( d \)) and \( \alpha \)-particles, we have:
- The mass of the deuteron: \( M_d = 2000 \, \text{MeV}/c^2 \)
- The mass of the \( \alpha \)-particle: \( M_\alpha = 4000 \, \text{MeV}/c^2 \)
- The kinetic energy of the deuteron: \( E_d = 10 \, \text{MeV} \)
- The kinetic energy of the \( \alpha \)-particle: \( E_\alpha = 20 \, \text{MeV} \)
Given that the charge \( q \) and magnetic field \( B \) are the same for both particles, the ratio of the radii is:
\[ \frac{r_\alpha}{r_d} = \frac{\sqrt{2M_\alpha E_\alpha}}{\sqrt{2M_d E_d}} = \sqrt{\frac{M_\alpha E_\alpha}{M_d E_d}} \]
Substituting the given values:
\[ \frac{r_\alpha}{r_d} = \sqrt{\frac{4000 \times 20}{2000 \times 10}} = \sqrt{\frac{80000}{20000}} = \sqrt{4} = 2 \]
Thus, the correct answer is: \( \boxed{2} \)
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