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:15.95 cm

Step-by-step explanation:

Given

Energy=300 kev

radius of Proton=16 cm

mass of alpha particle
`=6.64* 10^(-27) kg`

mass of proton
`=1.67* 10^(-27) kg`

charge on alpha particle is twice of proton

radius of Proton is given by

`r=(mv)/(|q|B)`

and Kinetic energy
`K=(P^2)/(2m)`

where P=momentum

`P=√(2Km)`

`r=(√(2km))/(qB)`---1

Radius for Alpha particle is

`r_(alpha)=\frac{\sqrt{2k\cdot m_(alpha)}}{2qB}`-----2

Divide 1 & 2 we get

`(r)/(r_(alpha))=(√(m))/(q)* \frac{2q}{\sqrt{m_(alpha)}}`

`r_(alpha)=\sqrt{(6.64* 10^(-27))/(1.67* 10^(-27))}* 0.5`

`r_(alpha)=0.997* 16`

`r_(alpha)=15.95 cm`

Answer 2

16 cm

Step-by-step explanation:

For protons:

Energy, E = 300 keV

radius of orbit, r1 = 16 cm

the relation for the energy and velocity is given by

`E = (1)/(2)mv^(2)`

So,
`v = \sqrt{(2E)/(m)}` .... (1)

Now,

`r = (mv)/(Bq)`

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

`r = (√(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

`(r_(1))/(r_(2))=(q_(2))/(q_(1))* \sqrt{(m_(1))/(m_(2))}`

q1 = q

q2 = 2q

m1 = m

m2 = 4m

By substituting the values

`(r_(1))/(r_(2))=\frac{2q}}{q}}* \sqrt{\frac{m}}{4m}}=1`

So, r2 = r1 = 16 cm

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