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`        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.`
5 years ago

```							Sol. V = 500 V B = 20 mT = (2 × 10^–3) T
E = V/d = 500/d ⇒ F = q500/d ⇒ a = q500/dm
⇒ u^2 = 2ad = 2 * q500/dm * d ⇒ u^2 = 100 * q/m ⇒ u √1000 * q/m
r base 1 = m base 1√1000 * q base 1/q base 1√m base 1B =  √m base 1√1000/√q base 1 B = √57 * 1.6 * 10^-27 * 10^3/√1.6 * 10^-19 * 2 * 10^-3 = 1.19 * 10^-2m = 119 cm
r base 1 = m base 2√1000 * q base 2/q base 2√m base 2B = √m base 2√1000/√q base 2 B = √1000 * 58 * 1.6 * 10^-27/√1.6 * 10^-19 *20 * 10^-3 = 1.20 * 10^-2 m = 120 cm

```
5 years ago
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