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+### Statement
+
+$10.1.9.$ [Insert the problem statement]
+
+### Solution
+
+### Statement
+
+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:
+$$D_{1\max} < D_{2\min}$$
+
+Substituting:
+$$\frac{2}{B}\sqrt{\frac{2m_1(V_0 + \Delta V)}{q}} < \frac{2}{B}\sqrt{\frac{2m_2(V_0 - \Delta V)}{q}}$$
+$$\sqrt{m_1(V_0 + \Delta V)} < \sqrt{m_2(V_0 - \Delta V)}$$
+
+After algebraic transformations, we get:
+$$\frac{\Delta V}{V_0} < \frac{m_2 - m_1}{m_1 + m_2}$$
+
+Substituting the masses of the potassium isotopes:
+$$\frac{\Delta V}{V_0} < \frac{41 - 39}{41 + 39} = \frac{2}{80} = 0.025$$
+
+#### Answer
+
+$$\frac{\Delta V}{V_0} < 0.025$$
+
+#### Answer
+
+[Insert a concise answer or boxed result]