Question

Singly charged uranium-238 ions are accelerated through a potential difference of 2.20 kV and enter a uniform magnetic field of 1.90 T directed perpendicular to their velocities. Determine the radius of their circular path.

Answer 1

r = 0.0548 m

Step-by-step explanation:

Given that,

Singly charged uranium-238 ions are accelerated through a potential difference of 2.20 kV and enter a uniform magnetic field of 1.90 T directed perpendicular to their velocities.

We need to find the radius of their circular path. The formula for the radius of path is given by :

`r=(1)/(B)\sqrt{(2mV)/(q)}`

m is mass of Singly charged uranium-238 ion,
`m=3.95* 10^(-25)\ kg`

q is charge

So,

`r=(1)/(1.9)* \sqrt{(2* 3.95* 10^(-25)* 2.2* 10^3)/(1.6* 10^(-19))}\\\\r=0.0548\ m`

So, the radius of their circular path is equal to 0.0548 m.