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 that 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 speed of the chlorine ions as they enter the magnetic field region

Answer 1

Step-by-step explanation:

From the question we are told that

The mass of chlorine ion is
`m_c  =  35u  =  35 * 1.66*10^(-27) =  5.81*10^(-26)\ kg`

The charge is
`q =  5e   = 5 *  1.60 *10^(-19) = 8.0*10^(-19)\ C`

The potential difference is
`V  =  250 kV  =  250*10^(3)  \  V`

The radius of curvature of the path is
`r = 3.5 \ m`

Gnerally the magnetic force will cause the speed of the chlorine ions to change from 0 m/s to
`v_y` m/s along the y -axis but will not affect the velocity along the x-axis

Generally according the law of energy conservation

`K =  PE`

Here K is the kinetic energy of the of the chlorine ions which is mathematically represented as

`K  =  (1)/(2) mv^2`

And PE is electric potential energy which is mathematically represented as

`PE  =   Q *  V`

So

`(1)/(2) mv^2 =  Q *  V`

=>
`(1)/(2) *  5.81*10^(-26)  *  v^2 =  8.0*10^(-19) *  250*10^(3)`

=>
`v =  sqrt{6.8847 *10^(12)}`

=>
`v = 2.634 *10^(6) \ m/s`