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| $10.1.9.$ [Insert the problem statement] |
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| 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? |
| 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? |
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| ### Solution |
| ### Solution |
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| When an ion passes through the accelerating voltage $V$, it acquires kinetic energy: |
| When an ion passes through the accelerating voltage $V$, it acquires kinetic energy: |
| $$\frac{mv^2}{2} = qV \implies v = \sqrt{\frac{2qV}{m}}$$ | | $$\frac{mv^2}{2} = qV \implies v = \sqrt{\frac{2qV}{m}}$$ |
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| In the magnetic field, the Lorentz force acts, which is centripetal: | | In the magnetic field, the Lorentz force acts, which is centripetal: |
| $$qvB = \frac{mv^2}{R} \implies R = \frac{mv}{qB}$$ | | $$qvB = \frac{mv^2}{R} \implies R = \frac{mv}{qB}$$ |
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| Substitute the velocity $v$: | | Substitute the velocity $v$: |
| $$R = \frac{m}{qB}\sqrt{\frac{2qV}{m}} = \frac{1}{B}\sqrt{\frac{2mV}{q}}$$ | | $$R = \frac{m}{qB}\sqrt{\frac{2qV}{m}} = \frac{1}{B}\sqrt{\frac{2mV}{q}}$$ |
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| The distance from the entrance to the field to the trace on the plate is the diameter $D = 2R$. | | The distance from the entrance to the field to the trace on the plate is the diameter $D = 2R$. |
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| 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: | | 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: |
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| The light isotope ($m_1 = 39$): its band ends where the voltage is maximum: | | The light isotope ($m_1 = 39$): its band ends where the voltage is maximum: |
| $$D_{1\max} = \frac{2}{B}\sqrt{\frac{2m_1(V_0 + \Delta V)}{q}}$$ | | $$D_{1\max} = \frac{2}{B}\sqrt{\frac{2m_1(V_0 + \Delta V)}{q}}$$ |
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| The heavy isotope ($m_2 = 41$): its band begins where the voltage is minimum: | | The heavy isotope ($m_2 = 41$): its band begins where the voltage is minimum: |
| $$D_{2\min} = \frac{2}{B}\sqrt{\frac{2m_2(V_0 - \Delta V)}{q}}$$ | | $$D_{2\min} = \frac{2}{B}\sqrt{\frac{2m_2(V_0 - \Delta V)}{q}}$$ |
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| 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: | | 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: |
| $$D_{1\max} < D_{2\min}$$ | | $$D_{1\max} < D_{2\min}$$ |
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| Substituting: | | Substituting: |
| $$\frac{2}{B}\sqrt{\frac{2m_1(V_0 + \Delta V)}{q}} < \frac{2}{B}\sqrt{\frac{2m_2(V_0 - \Delta V)}{q}}$$ | | $$\frac{2}{B}\sqrt{\frac{2m_1(V_0 + \Delta V)}{q}} < \frac{2}{B}\sqrt{\frac{2m_2(V_0 - \Delta V)}{q}}$$ |
| $$\sqrt{m_1(V_0 + \Delta V)} < \sqrt{m_2(V_0 - \Delta V)}$$ | | $$\sqrt{m_1(V_0 + \Delta V)} < \sqrt{m_2(V_0 - \Delta V)}$$ |
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| After algebraic transformations, we get: | | After algebraic transformations, we get: |
| $$\frac{\Delta V}{V_0} < \frac{m_2 - m_1}{m_1 + m_2}$$ | | $$\frac{\Delta V}{V_0} < \frac{m_2 - m_1}{m_1 + m_2}$$ |
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| Substituting the masses of the potassium isotopes: | | Substituting the masses of the potassium isotopes: |
| $$\frac{\Delta V}{V_0} < \frac{41 - 39}{41 + 39} = \frac{2}{80} = 0.025$$ | | $$\frac{\Delta V}{V_0} < \frac{41 - 39}{41 + 39} = \frac{2}{80} = 0.025$$ |
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