Question

In a mass spectrometer chlorine ions of mass 35u and charge +5e are emitted from a source and accelerated through a potential difference of 250 kV. They then enter a region with a magnetic field which is perpendicular to their original direction of motion. The chlorine ions exit the spectrometer after being bent along a path with radius of curvature 3.5 m. What is the value of the magnetic field? (u = 1.66

Answer 1

Magnetic field will be equal to 0.269 T

Step-by-step explanation:

We have given mass of chlorine ion is 35u

As we know that 1 u =
`=1.66* 10^(-27)kg`

Radius of circular path is given r = 3.5 m

So mass of chlorine ion
`=35* 1.66* 10^(-27)kg=58.1* 10^(-27)kg`

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

Potential difference V = 250 KV

From conservation of energy

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

So
`(1)/(2)* 58.1* 10^(-27)* v^2=1.6* 10^(-19)* 250000`

`v=2.6* 10^6m/sec`

We know that radius is equal to
`r=(mv)/(qB)`

So
`3.5=(58.1* 10^(-27)* 2.6* 10^6)/(1.6* 10^(-19)* B)`

`B=0.269Tesla`