Fe⁺ ions are accelerated through a potential difference of 500 V and are injected normally into a homogeneous magnetic field B of strength 20.0 mT. Find the radius of the circular paths followed by the isotopes with mass numbers 57 and 58. Take the mass of an ion = A (1.6 × 10⁻²⁷) kg, where A is the mass number.

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Solution

Given:
Potential difference through which the Fe⁺ ions are accelerated, V = 500 V
Strength of the homogeneous magnetic field, B = 20.0 mT = 20 × 10⁻³ T
Mass numbers of the two isotopes are 57 and 58.
Mass of an ion = A (1.6 × 10⁻²⁷) kg
We know that the radius of the circular path described by a particle in a magnetic field,
$r = \frac{m v}{q B} For isotope 1, r_{1} = \frac{m_{1} v_{1}}{q B} For isotope 2, r_{2} = \frac{m_{2} v_{2}}{q B} \Rightarrow \frac{r_{1}}{r_{2}} = \frac{m_{1} v_{1}}{m_{2} v_{2}}$

As both the isotopes are accelerated via the same potential V, the K.E gained by the two particles will be same.
$q V = \frac{1}{2} m_{1} {v_{1}}^{2} = \frac{1}{2} m_{2} {v_{2}}^{2} \frac{m_{1}}{m_{2}} = \frac{{v_{1}}^{2}}{{v_{2}}^{2}} \Rightarrow \frac{r_{1}}{r_{2}} = (\frac{m_{1}}{m_{2}})^{\frac{3}{2}}$

$Also, r_{1} = \frac{m v_{1}}{q B} = \frac{m_{1} \sqrt{\frac{2 q V}{m_{1}}}}{q B} = \frac{1}{B} \sqrt{\frac{2 m_{1} V}{q}}$
$= \frac{\sqrt{1000 \times 57 \times 1.6 \times 10^{- 27}}}{\sqrt{1.6 \times 10^{- 19}} \times 20 \times 10^{- 3}} = 1.19 \times 10^{- 2} m = 119 cm$
For the second isotope:
$As \frac{r_{1}}{r_{2}} = (\frac{m_{1}}{m_{2}})^{\frac{3}{2}}, r_{2} = (\frac{m_{2}}{m_{1}})^{\frac{3}{2}} r_{1} = (\frac{58}{57})^{\frac{3}{2}} \times 119 cm = 120 cm$

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