Question: A mass spectrograph is used to measure the masses of ions, or to separate ions of different masses. In one design for such an instrument, ions with mass $m$ and charge $q$ are accelerated through a potential difference $V$. They then enter a uniform magnetic field that is perpendicular to their velocity, and are deflected in a semicircular path of radius $R$. A detector measures where the ions complete the semicircle and from this it is easy to calculate $R$. Suppose the beam consists of a mixture of $^{12}C$ and $^{14}C$ ions. If $V$ and $B$ have the same values as in part B, calculate the separation of these two isotopes at the detector. Express your answer with the appropriate units.
1. **Problem Statement:** We have ions of two isotopes, $^{12}C$ and $^{14}C$, accelerated through the same potential difference $V$ and entering a magnetic field $B$. They follow semicircular paths with radii $R_{12}$ and $R_{14}$ respectively. We want to find the separation between these two isotopes at the detector, which is the difference in their radii $\Delta R = R_{14} - R_{12}$.
2. **Formula for radius in a mass spectrograph:** The radius $R$ of the semicircular path of an ion with mass $m$ and charge $q$ moving in a magnetic field $B$ after acceleration through potential difference $V$ is given by:
$$ R = \frac{mv}{qB} $$
where $v$ is the velocity of the ion after acceleration.
3. **Relating velocity to potential difference:** The kinetic energy gained by the ion is equal to the electric potential energy:
$$ \frac{1}{2}mv^2 = qV $$
Solving for $v$:
$$ v = \sqrt{\frac{2qV}{m}} $$
4. **Substitute $v$ into $R$ formula:**
$$ R = \frac{m}{qB} \times \sqrt{\frac{2qV}{m}} = \frac{m}{qB} \times \sqrt{\frac{2qV}{m}} = \frac{\sqrt{2mV/q}}{B} $$
Simplify:
$$ R = \frac{\sqrt{2mV}}{qB} $$
5. **Calculate radii for $^{12}C$ and $^{14}C$:** Since $q$, $V$, and $B$ are the same for both isotopes, the radius depends on $\sqrt{m}$. Let $m_{12}$ and $m_{14}$ be the masses of $^{12}C$ and $^{14}C$ respectively.
$$ R_{12} = \frac{\sqrt{2m_{12}V}}{qB} $$
$$ R_{14} = \frac{\sqrt{2m_{14}V}}{qB} $$
6. **Find the separation $\Delta R$:**
$$ \Delta R = R_{14} - R_{12} = \frac{\sqrt{2V}}{qB} (\sqrt{m_{14}} - \sqrt{m_{12}}) $$
7. **Interpretation:** The separation depends on the difference of the square roots of the masses of the isotopes.
8. **Final answer:**
$$ \boxed{\Delta R = \frac{\sqrt{2V}}{qB} (\sqrt{m_{14}} - \sqrt{m_{12}})} $$
This formula gives the separation of the $^{12}C$ and $^{14}C$ ion beams at the detector in terms of known quantities $V$, $q$, $B$, and the masses $m_{12}$ and $m_{14}$.