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 $1.87 \times 10^{-5}$ MV.

Explanation:

The given data is as follows.

    $m_{1}$ = 35 amu

              = $35 \times 1.66 \times 10^{-27}$

              =  kg

     = 37 amu

          = $37 \times 1.66 \times 10^{-27}$

          =  kg

It is known that,

    work done on charged particle = gain in kinetic energy

So,      

               v = $\sqrt{(2 \times q \times \Delta_{V/m})}$

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

Hence,     F = $q \times v \times B (sin(90))$

        $\frac{m \times \frac{v^{2}}{r} = q \times v \times B$

                    r = $\frac{m \times v}{B \times q}$

And,      

             $r_{2} = \frac{m_{2} \times v_{2}}{(B \times q)}$

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

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

 $0.7 \times 10^{-2} = \sqrt{(\Delta_{V})} \times (\sqrt(2 \times \frac{m_{2}}{q}) - \sqrt{\frac{(2 \times \frac{m_{1}}{q}))}{B}}$

$\sqrt{(\Delta_V)} = 0.7 \times 10^{-2} \times \frac{B}{(\sqrt{(2 \times \frac{m_{2}{q})} - \sqrt{(\frac{2 \times m_{1}}{q}))}$

= $\frac{0.7 \times 10^{-2} \times 1.2}{(\sqrt{(2 \times \frac{6.142 \times 10^{-26}}{(1.6 \times 10^{-19}})}) - \sqrt{(2 \times \frac{5.81 \times 10^{-26}}{(1.6 \times 10^{-19}))}$

               = 350 volts

    $\Delta_V$ = $\sqrt{350}$

                   = 18.7 volts  

                  = $1.87 \times 10^{-5}$ MV

Thus, we can conclude that the minimum value of the potential difference through which these ions must be accelerated is $1.87 \times 10^{-5}$ MV.