10.1.9. In a device for determining the isotopic composition, potassium ions $^{39}K^+$ and $^{41}K^+$ are first accelerated in an electric field and then enter a uniform magnetic field with induction $B$, perpendicular to their direction of motion. During the experiment, due to the imperfection of the apparatus, the accelerating voltage fluctuates around its average value by an amount of $\pm\Delta V$. With what relative error $\frac{\Delta V}{V_0}$ must the value of the accelerating voltage be kept constant so that the traces of the potassium isotope beams on the photographic plate $\Phi$ do not overlap?
Solution
When an ion passes through the accelerating voltage $V$, it acquires kinetic energy: $$\frac{mv^2}{2} = qV \implies v = \sqrt{\frac{2qV}{m}}$$
In the magnetic field, the Lorentz force acts, which is centripetal: $$qvB = \frac{mv^2}{R} \implies R = \frac{mv}{qB}$$
Substitute the velocity $v$: $$R = \frac{m}{qB}\sqrt{\frac{2qV}{m}} = \frac{1}{B}\sqrt{\frac{2mV}{q}}$$
The distance from the entrance to the field to the trace on the plate is the diameter $D = 2R$.
Because the voltage $V$ varies from $V_{\min} = (V_0 - \Delta V)$ to $V_{\max} = (V_0 + \Delta V)$, instead of a thin line, each isotope leaves a band on the plate:
The light isotope ($m_1 = 39$): its band ends where the voltage is maximum: $$D_{1\max} = \frac{2}{B}\sqrt{\frac{2m_1(V_0 + \Delta V)}{q}}$$
The heavy isotope ($m_2 = 41$): its band begins where the voltage is minimum: $$D_{2\min} = \frac{2}{B}\sqrt{\frac{2m_2(V_0 - \Delta V)}{q}}$$
For the beams not to overlap, the right edge of the light isotope band must be to the left of the left edge of the heavy isotope band: $$D_{1\max} < D_{2\min}$$
