After being accelerated through a potential difference of 500 V , Fe ^+ ions are injected normally into a homogeneous magnetic field B having a strength of 20.0 mT . Determine the radius of the circular paths described by the isotopes with mass numbers 57 and 58 . Consider the mass of an ion to be A(1.6 10⁻²⁷) kg , where A represents the mass number.

Step-by-step solution

The kinetic energy acquired by an ion accelerated through a potential difference V is K = qV . The radius of the circular path described by an ion in a uniform magnetic field is given by R = mv qB = 2mK qB = 1 B 2mV q . Given values are: V = 500 V B = 20.0 10⁻³ T q = 1.6 10⁻¹⁹ C m = A 1.6 10⁻²⁷ kg Substituting these values into the formula for radius: R = 1 20.0 10⁻³ 2 A 1.6 10⁻²⁷ 500 1.6 10⁻¹⁹ R = 50 A 10⁻⁵ m R = 0.05 10A m = 5 10A cm For the isotope with mass number A = 57 : R₅₇ = 5 570 5 23.87 cm 119.35 cm 119 c