Answer:
[tex]r_\alpha=16cm[/tex]
Explanation:
The radius of the circumference described by a particle in a cyclotron is given by:
[tex]r=\frac{mv}{qB}(1)[/tex]
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:
[tex]K=\frac{mv^2}{2}=\frac{m^2v^2}{2m}\\[/tex]
Solving this for mv:
[tex]mv=\sqrt{2mK}(2)[/tex]
Replacing (2) in (1):
[tex]r=\frac{\sqrt{2mK}}{qB}[/tex]
For protons, we have:
[tex]r_p=\frac{\sqrt{2m_pK}}{eB}(3)[/tex]
For alpha particles, we have:
[tex]r_\alpha=\frac{\sqrt{2m_\alpha K}}{(2e)B}(4)[/tex]
Dividing (4) in (3):
[tex]\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[/tex]
The orbital radius of the alpha particle in the constant magnetic field is 16 cm.
The given parameters;
The orbital radius of the alpha particles is calculated as follows;
[tex]\frac{mv^2}{r} = qvB\\\\r = \frac{mv}{qB} \ --(1)\\\\K.E = \frac{1}{2} mv^2\\\\mv^2 = 2K.E\\\\m(mv^2) = 2K.E\ m\\\\mv = \sqrt{2K.E \ m} \ --(2)\\\\r = \frac{\sqrt{2K.E \ m}}{qB}[/tex]
The radius of the proton;
[tex]r_p = \frac{\sqrt{2K.E \ m_p}}{eB}[/tex]
The radius of the alpha - particle;
[tex]r_{\alpha} = \frac{\sqrt{2K.E \ m_{\alpha}}}{2eB}[/tex]
The orbital radius of the alpha particle is calculated as;
[tex]\frac{r_{\alpha}}{r_p} = \frac{ \frac{\sqrt{2K.E \ m_{\alpha}}}{2eB}}{ \frac{\sqrt{2K.E \ m_p}}{eB}} \\\\\frac{r_{\alpha}}{r_p} = \frac{1}{2} \sqrt{\frac{m_{\alpha}}{m_p} } \\\\\frac{r_{\alpha}}{r_p} = \frac{1}{2} \sqrt{\frac{6.64\times 10^{-27}}{1.67\times 10^{-27}} }\\\\\frac{r_{\alpha}}{r_p} = \frac{1}{2} (2)\\\\\frac{r_{\alpha}}{r_p} = 1\\\\r_{\alpha} = 1 \times r_p\\\\r_{\alpha} = 1 \times 16 \ cm\\\\r_{\alpha} = 16 \ cm[/tex]
Thus, the orbital radius of the alpha particle is 16 cm.
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