Question

Chlorine has two stable isotopes, 35Cl and 37Cl. Chlorine gas which consists of singly ionized ions is to be separated into its isotopic components using a mass spectrometer. The magnetic field strength in the spectrometer is 1.2 T. What is the minimum value of the potential difference through which these ions must be acceler- ated so that the separation between them, after they complete their semicircular path, is 1.4 cm?

Answer 1

Answer: The minimum value of the potential difference through which these ions must be accelerated is
`1.87 * 10^(-5)` MV.

Step-by-step explanation:

The given data is as follows.

`m_(1)` = 35 amu

=
`35 * 1.66 * 10^(-27)`

= kg

= 37 amu

=
`37 * 1.66 * 10^(-27)`

= kg

It is known that,

work done on charged particle = gain in kinetic energy

So,

v =
`\sqrt{(2 * q * \Delta_(V/m))}`

A centripetal force is experienced when charged particles enter a uniform magnetic field.

Hence, F =
`q * v * B (sin(90))`

r =
`(m * v)/(B * q)`

And,

`r_(2) = (m_(2) * v_(2))/((B * q))`

On completion of semi circle, the distance between two paths =

`0.7 * 10^(-2) = (\sqrt(2 * m_(2) * delta_(V/q)) -\sqrt{(2 * m_(1) * (\Delta_(V/q))/(B)}`

`0.7 * 10^(-2) = \sqrt{(\Delta_(V))} * (\sqrt(2 * (m_(2))/(q)) - \sqrt{((2 * (m_(1))/(q))))/(B)}`

=

= 350 volts

`\Delta_V` =
`√(350)`

= 18.7 volts

=
`1.87 * 10^(-5)` MV

Thus, we can conclude that the minimum value of the potential difference through which these ions must be accelerated is
`1.87 * 10^(-5)` MV.