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Chapter 27: Problem 60

A mass spectrograph is used to measure the masses of ions, or to separate ions of different masses (see Section 27.5). 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 they 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 \(R\). (a) Derive the equation for calculating the mass of the ion from measurements of \(B\), \(V\), \(R\), and \(q\). (b) What potential difference \(V\) is needed so that singly ionized \(^{12}\)C atoms will have \(R =\) 50.0 cm in a 0.150-T magnetic field? (c) 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. Do you think that this beam separation is sufficient for the two ions to be distinguished? (Make the assumption described in Problem 27.59 for the masses of the ions.)

Short Answer

Expert verified
Formula for mass: \(m = \frac{qB^2R^2}{2V}\). (b) Potential \(V \approx 883\) V. (c) Separation = 2 cm, sufficient to distinguish isotopes.

Step by step solution

01

Understanding the ion motion in a magnetic field

The ions, once accelerated by a potential difference, enter the magnetic field and move in a semicircular path. The magnetic force provides the centripetal force necessary for this circular motion. The force can be described by: \[ \frac{mv^2}{R} = qvB \]
02

Relating ion velocity to the potential difference

The voltage through which the ion is accelerated gives it kinetic energy: \[ \frac{1}{2}mv^2 = qV \]. Solving for \(v\), we find: \[ v = \sqrt{\frac{2qV}{m}} \].
03

Substitute velocity in the force equation

We substitute the velocity \(v = \sqrt{\frac{2qV}{m}}\) into the magnetic force equation: \[ \frac{m(\sqrt{\frac{2qV}{m}})^2}{R} = q(\sqrt{\frac{2qV}{m}})B \]. Simplifying gives us: \[ m = \frac{qB^2R^2}{2V} \].
04

Calculate the potential difference for 12C

For \(^{12}\text{C}\) with \(R = 0.50 \text{ m}\), \(B = 0.150 \text{ T}\), and \(q = 1.6 \times 10^{-19}\text{ C (singly ionized)}\), use the equation: \[ V = \frac{qB^2R^2}{2m} \]. The atomic mass is \(12 \times 1.66 \times 10^{-27}\text{ kg}\). Solving gives \[ V \approx 883\text{ V}\].
05

Calculate path radius for 14C and separation

For \(^{14}C\), with \(m = 14 \times 1.66 \times 10^{-27}\text{ kg}\), using the same \(V\) calculated in Step 4, compute \(R\): \[ R = \frac{v \cdot m}{qB} \]. The difference in radii for \(^{12}C\) and \(^{14}C\) is the separation. This results in \( \Delta R \approx 0.02\text{ m} \), or 2 cm.
06

Evaluate separation sufficiency

With a separation of 2 cm, given practical constraints and the precision of measurement instruments, this separation is adequate to distinguish between \(^{12}C\) and \(^{14}C\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ion Separation
In a mass spectrograph, ions are separated based on their mass-to-charge ratio, allowing for the identification and measurement of different ions. When ions enter the device, they are subjected to a magnetic field that causes them to follow a path determined by their mass and charge. Heavier ions or those with higher charges will follow a different path than lighter ones, leading to their separation by the spectrograph.
  • This separation allows for precise identification of ions in samples.
  • The separation typically results in different arrival points at the detector based on the ion's mass-to-charge ratio.
Understanding ion separation is crucial for interpreting the results and ensuring accuracy in experiments involving ionized particles.
Centripetal Force
When ions move through a magnetic field in a mass spectrograph, they experience a centripetal force. This force is what keeps ions moving in a circular, or more practically, a semicircular path inside the device. The magnetic force is crucial here, acting as the centripetal force necessary for ion motion. The equation that relates mass (\(m\)), velocity (\(v\)), radius (\(R\)), charge (\(q\)), and magnetic field (\(B\)) is: \[ \frac{m v^2}{R} = q v B \].
  • The centripetal force here ensures the ions curve through the device rather than traveling in a straight line.
  • This force is a balance between the ion's inertia and the magnetic field's influence.
It is essential for maintaining the desired path and enabling accurate measurement of ion properties.
Magnetic Field
The magnetic field in a mass spectrograph is a key component that influences the path of ions. It is perpendicular to the motion of the ions, causing them to deflect and move in a circular path. This deflection is due to the force exerted by the field, which acts as the centripetal force on the moving ions. The strength of the magnetic field further determines how much the ions will curve.
  • A stronger magnetic field results in a sharper curvature, affecting the path radius.
  • This key factor affects where the ions ultimately land in the detector region.
By controlling the magnetic field, researchers can effectively manipulate ion paths and achieve desired results in measurements.
Potential Difference
The potential difference through which ions are accelerated in a mass spectrograph provides them with kinetic energy. This energy propels the ions into the magnetic field, influencing their velocity. The relationship between potential difference (\(V\)), charge (\(q\)), and energy is given by: \[ \frac{1}{2} m v^2 = q V \], which further allows us to solve for velocity (\(v = \sqrt{\frac{2qV}{m}}\)).
  • Alterations in potential difference can change the speed of ion movement, thereby impacting their path in the magnetic field.
  • This parameter is crucial for determining the radius of the ion's path and thus its position at the detector.
The careful manipulation and calculation of potential difference are vital for precision in experiments.
Isotope Separation
Isotope separation in a mass spectrograph involves distinguishing between ions of different isotopes, such as carbon-12 and carbon-14. Despite having the same chemical properties, isotopes differ slightly in mass, which affects their path in the spectrograph. This subtle difference allows the separation to occur, given precise control over the experimental conditions.
  • The device uses the differences in mass-to-charge ratios to separate isotopes along different paths.
  • Seeing these paths lets scientists measure the isotopic composition of a sample effectively.
This separation is often critical in fields like geology and archaeology, where isotope ratios are used for dating and environmental studies.

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Most popular questions from this chapter

In the Bainbridge mass spectrometer (see Fig. 27.24), the magnetic-field magnitude in the velocity selector is 0.510 T, and ions having a speed of 1.82 \(\times\) 10\(^6\) m/s pass through undeflected. (a) What is the electric-field magnitude in the velocity selector? (b) If the separation of the plates is 5.20 mm, what is the potential difference between the plates?

A conducting bar with mass m and length \(L\) slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current \(I\) in the rails and bar, and a constant, uniform, vertical magnetic field \(\overrightarrow{B}\) fills the region between the rails (\(\textbf{Fig. P27.65}\)). (a) Find the magnitude and direction of the net force on the conducting bar. Ignore friction, air resistance, and electrical resistance. (b) If the bar has mass \(m\), find the distance \(d\) that the bar must move along the rails from rest to attain speed \(v\). (c) It has been suggested that rail guns based on this principle could accelerate payloads into earth orbit or beyond. Find the distance the bar must travel along the rails if it is to reach the escape speed for the earth (11.2 km/s). Let \(B =\) 0.80 T, \(I =\) 2.0 \(\times\) 10\(^3\) A, \(m =\) 25 kg, and \(L =\) 50 cm. For simplicity assume the net force on the object is equal to the magnetic force, as in parts (a) and (b), even though gravity plays an important role in an actual launch in space.

An alpha particle (a He nucleus, containing two protons and two neutrons and having a mass of 6.64 \(\times\) 10\(^{-27}\) kg) traveling horizontally at 35.6 km>s enters a uniform, vertical, 1.80-T magnetic field. (a) What is the diameter of the path followed by this alpha particle? (b) What effect does the magnetic field have on the speed of the particle? (c) What are the magnitude and direction of the acceleration of the alpha particle while it is in the magnetic field? (d) Explain why the speed of the particle does not change even though an unbalanced external force acts on it.

A particle with charge 7.26 \(\times\) 10\(^{-8}\) C is moving in a region where there is a uniform 0.650-T magnetic field in the +\(x\)-direction. At a particular instant, the velocity of the particle has components \(v_x =\) -1.68 \(\times\) 10\(^4\) m/s, \(v_y =\) -3.11 \(\times\) 104 m/s, and \(v_z =\) 5.85 \(\times\) 10\(^4\) m/s. What are the components of the force on the particle at this time?

A circular loop of wire with area \(A\) lies in the \(xy\)-plane. As viewed along the \(z\)-axis looking in the -\(z\)-direction toward the origin, a current \(I\) is circulating clockwise around the loop. The torque produced by an external magnetic field \(\overrightarrow{B}\) is given by \(\vec{\tau}\) = D(4\(\hat{\imath}\) - 3\(\hat{\jmath}\)), where \(D\) is a positive constant, and for this orientation of the loop the magnetic potential energy \(U = -\vec{\mu}\) \(\cdot\) \(\overrightarrow{B}\) is negative. The magnitude of the magnetic field is \(B_0 = 13D/IA\). (a) Determine the vector magnetic moment of the current loop. (b) Determine the components \(B_x\), \(B_y\), and \(B_z\) of \(\overrightarrow{B}\).
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