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 :

Answer:

Explanation:

From  the question we are told that

   The mass of chlorine ion is [tex]m_c  =  35u  =  35 * 1.66*10^{-27} =  5.81*10^{-26}\ kg[/tex]

    The charge is  [tex]q =  5e   = 5 *  1.60 *10^{-19} = 8.0*10^{-19}\ C[/tex]

   The potential difference is  [tex]V  =  250 kV  =  250*10^{3}  \  V[/tex]

    The radius of curvature of the path is  [tex]r = 3.5 \ m[/tex]

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

Generally according the law of energy conservation

          [tex]K =  PE[/tex]

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

      [tex]K  =  \frac{1}{2} mv^2[/tex]

And  PE is electric potential  energy which is mathematically represented as

       [tex]PE  =   Q *  V[/tex]

So

      [tex] \frac{1}{2} mv^2 =  Q *  V [/tex]

=>   [tex] \frac{1}{2} *  5.81*10^{-26}  *  v^2 =  8.0*10^{-19} *  250*10^{3} [/tex]

=>    [tex] v =  sqrt{6.8847 *10^{12}} [/tex]

=>    [tex] v = 2.634 *10^{6} \ m/s [/tex]

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