Question

In a cyclotron, the orbital radius of protons with energy 300 keV is 16.0 cm . You are redesigning the cyclotron to be used instead for alpha particles with energy 300 keV . An alpha particle has charge q=+2e and massm=6.64×10−27kg .
If the magnetic field isnt changed, what will be the orbital radius of the alpha particles?

Answer 1

Answer:

$r_\alpha=16cm$

Explanation:

The radius of the circumference described by a particle in a cyclotron is given by:

$r=\frac{mv}{qB}(1)$

m is the particle's mass, v is the speed of the particle, q is the particle's charge and B is the magnitude of the magnetic field.

Kinetic energy is defined as:

$K=\frac{mv^2}{2}=\frac{m^2v^2}{2m}$

Solving this for mv:

$mv=\sqrt{2mK}(2)$

Replacing (2) in (1):

$r=\frac{\sqrt{2mK}}{qB}$

For protons, we have:

$r_p=\frac{\sqrt{2m_pK}}{eB}(3)$

For alpha particles, we have:

$r_\alpha=\frac{\sqrt{2m_\alpha K}}{(2e)B}(4)$

Dividing (4) in (3):

$\frac{r_\alpha}{r_p}=\frac{\frac{\sqrt{2m_\alpha K}}{(2e)B}}{\frac{\sqrt{2m_p K}}{(e)B}}\\r_\alpha=\frac{r_p}{2}\sqrt{\frac{m_\alpha}{m_p}}\\r_\alpha=\frac{16cm}{2}\sqrt{\frac{6.64*10^{-27}kg}{1.67*10^{-27}kg}}\\r_\alpha=\frac{16cm}{2}(\sqrt{3.98})\\\\r_\alpha=\frac{16cm}{2}(2)\\r_\alpha=16cm$