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Mixed \(\mathrm{He}^{+}\)and \(\mathrm{O}^{2+}\) ions (mass of \(\mathrm{He}^{+}=4\) amu and that of \(\mathrm{O}^{2+}=16 \mathrm{amu}\) ) beam passes a region of constant perpendicular magnetic field. It kinetic energy of all the ions is same, then (a) \(\mathrm{He}^{+}\)ions will be deflected more than those of \(\mathrm{O}^{2+}\) (b) \(\mathrm{He}^{+}\)ions will be deflected less than that of \(\mathrm{O}^{2+}\) (c) all the ions will be deflected equally (d) no ions will be deflected

Short Answer

Expert verified
(c) all the ions will be deflected equally

Step by step solution

01

Understanding ion deflection in a magnetic field

Ions moving through a magnetic field experience a force called the Lorentz force, which causes them to move in a circular path. This force is given by the equation \( F = qvB \sin(\theta) \), where \( q \) is the charge, \( v \) is the velocity, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the velocity and the magnetic field. For a perpendicular field, \( \sin(\theta) = 1 \). The radius \( r \) of the circular path is calculated using \( r = \frac{mv}{qB} \), where \( m \) is the mass of the ion.
02

Relate kinetic energy to velocity

The kinetic energy \( K \) of an ion is given by \( K = \frac{1}{2}mv^2 \). If all ions have the same kinetic energy, we can express \( v \) in terms of \( K \) and \( m \) as \( v = \sqrt{\frac{2K}{m}} \).
03

Substituting velocity into radius equation

Substitute the expression for \( v \) from Step 2 into the equation for the radius of the circular path: \( r = \frac{m\sqrt{\frac{2K}{m}}}{qB} \). This simplifies to \( r = \frac{\sqrt{2Km}}{qB} \).
04

Comparing deflections for He鈦 and O虏鈦 ions

For \( \mathrm{He}^{+} \) with mass 4 amu and charge \( +1 \), and \( \mathrm{O}^{2+} \) with mass 16 amu and charge \( +2 \), calculate the radius \( r \) for each. \( r \) is directly proportional to \( \sqrt{m} \) and inversely proportional to \( q \), so: \( r_{\mathrm{He}^{+}} = \frac{\sqrt{2K \cdot 4}}{1B} \) and \( r_{\mathrm{O}^{2+}} = \frac{\sqrt{2K \cdot 16}}{2B} \). Both reduce proportionally with the same kinetic energy.
05

Calculating radii for simplification

Simplify the expressions for the radii: \( r_{\mathrm{He}^{+}} = \frac{\sqrt{8K}}{B} \) and \( r_{\mathrm{O}^{2+}} = \frac{\sqrt{32K}}{2B} = \frac{\sqrt{8K}}{B} \). Both have the same simplified expression.

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Key Concepts

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

Lorentz Force
Imagine ions flying through a cosmic dance floor, the magnetic field. This magnetic field exerts a special force on them known as the Lorentz force. It's like an invisible hand pushing them along a curved path instead of letting them go straight. Think of it as the force behind the curtains at a theatre, making the actors (ions) move in circles across the stage. The Lorentz force can be calculated by the formula \( F = qvB \sin(\theta) \), where:
  • \( q \) represents the charge of the ion
  • \( v \) is the velocity at which it moves
  • \( B \) signifies the strength of the magnetic field
  • \( \theta \) is the angle between their velocity and the magnetic field direction
However, when the magnetic field is perpendicular (\( \theta = 90^{\circ} \)), this force is at its maximum because \( \sin(\theta) = 1 \). In other words, the more the charge or the speed, or the stronger the field, the greater the force urging the ions to curve. It鈥檚 the playing field that decides how curved their paths will be.
Circular Motion in Magnetic Fields
When Lorentz force steps into the game, it converts every ion鈥檚 journey into a circular one. Each ion traces a circular path due to this force, a bit like how planets orbit around the sun. The radius of this path (let's say path radius \( r \)) can tell us a ton about their motion. It鈥檚 determined using the formula \( r = \frac{mv}{qB} \), where \( m \) is the mass of the ion.
  • Big mass (\( m \))? Larger circle.
  • Higher charge (\( q \))? Smaller circle.
  • Stronger magnet (\( B \))? Tighter loop.
Now, combining this with the kinetic energy from our previous section, we can substitute the expression for velocity into this formula to further simplify the radius calculation. This helps us understand how different ions such as \( \text{He}^+ \) and \( \text{O}^{2+} \) ions would trace different circular paths when zipping through the same magnetic field.
Kinetic Energy and Velocity Relationship
Kinetic energy is the energy of motion and dictates how fast an ion moves through the magnetic field. For any moving object, including ions, it is defined by the formula \( K = \frac{1}{2}mv^2 \), which ties together mass and speed. By rearranging this formula, we can express velocity \( v \) in terms of kinetic energy \( K \) and mass \( m \). The expression transforms to \( v = \sqrt{\frac{2K}{m}} \). This relationship indicates:
  • More energy (\( K \))? Faster ion.
  • Heavier ion (\( m \))? Slower speed, with the same energy.
When given equal kinetic energy, as in our exercise, lighter ions like \( \text{He}^+ \) move quicker than heavier ones like \( \text{O}^{2+} \) because of this inverse relationship with mass. By substituting this velocity expression back into our radius equation from the circular motion concept, we gain insight into the circular path鈥檚 radius for each type of ion and their respective deflection patterns.

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Most popular questions from this chapter

The ratio of molecular mass of two radioactive substances is \(3 / 2\) and the ratio of their decay constants is \(4 / 3\). Then, the ratio of their initial activity per mole will be (a) 2 (b) \(4 / 3\) (c) \(8 / 9\) (d) \(9 / 8\)

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