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alina1380 [7]
4 years ago
6

Can someone please help meee

Physics
1 answer:
Advocard [28]4 years ago
4 0

Answer:

I think your answer might be letter C

Explanation:

If it is wrong I am so sorry forgive me if it is wrong let me know please ok.

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In a cyclotron, the orbital radius of protons with energy 300 keV is 16.0 cm . You are redesigning the cyclotron to be used inst
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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:

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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

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