Question

Two ions of masses 4 amu and 16 amu have charges +2e and +3e respectively. These ions pass through the region of constant perpendicular magnetic field. The kinetic energy of both ions is same. Then :

• A. lighter ion will be deflected more than heavier ion
• B. lighter ion will be deflected less than heavier ion
• C. both ions will be deflected equally
• D. no ion will be deflected

Hint:

Deflection is inversely proportional to the radius of the circular path (\(r\)). Always express the radius in terms of the given quantities. In this case, since kinetic energy (\(K\)) is given, use the formula \(r = \frac{\sqrt{2mK}}{qB}\). Then analyze the proportionality \(r \propto \frac{\sqrt{m}}{q}\).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
Two ions with different masses and charges enter a uniform magnetic field with the same kinetic energy. We need to compare their deflection. Deflection is inversely related to the radius of the circular path they follow. A smaller radius means a larger deflection.
Step 2: Key Formula or Approach:
When a charged particle moves perpendicular to a magnetic field, it follows a circular path. The radius of this path is given by: \[ r = \frac{mv}{qB} \] where \(m\) is mass, \(v\) is velocity, \(q\) is charge, and \(B\) is the magnetic field strength.
The kinetic energy is \(K = \frac{1}{2}mv^2\). We can express momentum \(mv\) in terms of kinetic energy: \(mv = \sqrt{2mK}\).
Substituting this into the radius formula: \[ r = \frac{\sqrt{2mK}}{qB} \] Step 3: Detailed Explanation:
Let the lighter ion be ion 1 and the heavier ion be ion 2.
Given data:
Mass of lighter ion, \(m_1 = 4\) amu
Charge of lighter ion, \(q_1 = +2e\)
Mass of heavier ion, \(m_2 = 16\) amu
Charge of heavier ion, \(q_2 = +3e\)
Kinetic energy is the same for both: \(K_1 = K_2 = K\).
The magnetic field is also the same: \(B_1 = B_2 = B\).
Now we calculate the ratio of their radii. Since \(K\) and \(B\) are constant, the radius \(r\) is proportional to \(\frac{\sqrt{m}}{q}\).
Radius of the lighter ion's path: \[ r_1 \propto \frac{\sqrt{m_1}}{q_1} = \frac{\sqrt{4}}{2} = \frac{2}{2} = 1 \] Radius of the heavier ion's path: \[ r_2 \propto \frac{\sqrt{m_2}}{q_2} = \frac{\sqrt{16}}{3} = \frac{4}{3} \approx 1.33 \] Comparing the radii, we find that \(r_1<r_2\).
Since the radius of the path of the lighter ion (\(r_1\)) is smaller than that of the heavier ion (\(r_2\)), the lighter ion follows a more tightly curved path.
Step 4: Final Answer:
A more curved path means a greater deflection from the original direction of motion. Therefore, the lighter ion will be deflected more than the heavier ion.