Question

Suppose that an ion source in a mass spectrometer produces doubly ionized gold ions (Au2+), each with a mass of 3.27 × 10-25 kg. The ions are accelerated from rest through a potential difference of 1.90 kV. Then, a 0.540-T magnetic field causes the ions to follow a circular path. Determine the radius of the path.

Answer 1

Radius of the path is
`163.16* 10^(-6)m`

Step-by-step explanation:

It is given that gold is doubly ionized

Charge
`q=2e=2* 1.6* 10^(-19)=3.2* 10^(-19)C`

Mass of the each ion =
`=3.27* 10^(-25)kg`

So total mass m =
`=2* 3.27* 10^(-25)=6.54* 10^(-25)kg`

Potential difference V = 1.9 KV

Magnetic field B = 0.540 T

We know that
`(1)/(2)mv^2=qV`

`(1)/(2)* 6.54* 10^(-25)* v^2=3.2* 10^(-19)* 1900`

`v=43.11* 10^3m/sec`

We have to find the radius of the path

We know that radius of the path is given by

`r=(mv)/(qB)=(6.54* 10^(-25)* 43.11* 10^3)/(3.2* 10^(-16)* 0.540)=163.16* 10^(-6)m`