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? Express your answer to three significant figures and include the appropriate units.

Answer 1

Answer:

16 cm

Explanation:

For protons:

Energy, E = 300 keV

radius of orbit, r1 = 16 cm

the relation for the energy and velocity is given by

$E = \frac{1}{2}mv^{2}$

So, $v = \sqrt{\frac{2E}{m}}$   .... (1)

Now,

$r = \frac{mv}{Bq}$

Substitute the value of v from equation (1), we get

$r = \frac{\sqrt{2mE}}{Bq}$

Let the radius of the alpha particle is r2.

For proton

So, $r_{1} = \frac{\sqrt{2m_{1}E}}{Bq_{1}}$    ... (2)

Where, m1 is the mass of proton, q1 is the charge of proton

For alpha particle

So, $r_{2} = \frac{\sqrt{2m_{2}E}}{Bq_{2}}$    ... (3)

Where, m2 is the mass of alpha particle, q2 is the charge of alpha particle

Divide equation (2) by equation (3), we get

$\frac{r_{1}}{r_{2}}=\frac{q_{2}}{q_{1}}\times \sqrt{\frac{m_{1}}{m_{2}}}$

q1 = q

q2 = 2q

m1 = m

m2 = 4m

By substituting the values

$\frac{r_{1}}{r_{2}}=\frac{2q}}{q}}\times \sqrt{\frac{m}}{4m}}=1$

So, r2 = r1 = 16 cm

Thus, the radius of the alpha particle is 16 cm.

Answer 2

Answer:15.95 cm

Explanation:

Given

Energy=300 kev

radius of Proton=16 cm

mass of alpha particle$=6.64\times 10^{-27} kg$

mass of proton$=1.67\times 10^{-27} kg$

charge on alpha particle is twice of proton

radius of Proton is given by

$r=\frac{mv}{|q|B}$

and Kinetic energy $K=\frac{P^2}{2m}$

where P=momentum

$P=\sqrt{2Km}$

$r=\frac{\sqrt{2km}}{qB}$---1

Radius for Alpha particle is

$r_{alpha}=\frac{\sqrt{2k\cdot m_{alpha}}}{2qB}$-----2

Divide 1 & 2 we get

$\frac{r}{r_{alpha}}=\frac{\sqrt{m}}{q}\times \frac{2q}{\sqrt{m_{alpha}}}$

$r_{alpha}=\sqrt{\frac{6.64\times 10^{-27}}{1.67\times 10^{-27}}}\times 0.5$

$r_{alpha}=0.997\times 16$

$r_{alpha}=15.95 cm$