Fe+ ions are accelerated with P.d= 500V. and injected normally into a Magnet field of 20mT. Find the radius od the circular path followed by two isotopes of mass no. 57 and 58.???Mion= 1.6x 10^-27 Kg......

To determine the radius of the circular path followed by the Fe⁺ ions in a magnetic field, we can use the principles of electromagnetism and the motion of charged particles in magnetic fields. The key equations we will use involve the kinetic energy of the ions and the magnetic force acting on them.

Step 1: Calculate the Velocity of the Ions

When the Fe⁺ ions are accelerated through a potential difference (P.d) of 500 V, they gain kinetic energy equal to the work done on them by the electric field. The kinetic energy (KE) can be expressed as:

KE = qV

Where:

• q is the charge of the ion (for Fe⁺, q = +1.6 x 10^-19 C),
• V is the potential difference (500 V).

Substituting the values:

KE = (1.6 x 10^-19 C)(500 V) = 8.0 x 10^-17 J

The kinetic energy is also related to the mass and velocity of the ion:

KE = 0.5mv²

Setting the two expressions for kinetic energy equal gives:

0.5mv² = 8.0 x 10^-17 J

Solving for velocity (v), we have:

v = sqrt((2 * KE) / m)

Step 2: Calculate the Radius of the Circular Path

When a charged particle moves in a magnetic field, it experiences a magnetic force that acts as a centripetal force, causing it to move in a circular path. The magnetic force (F_B) is given by:

F_B = qvB

Where:

• B is the magnetic field strength (20 mT = 20 x 10^-3 T).

This magnetic force provides the centripetal force required to keep the ion in circular motion:

F_C = (mv²) / r

Setting the magnetic force equal to the centripetal force gives:

qvB = (mv²) / r

Rearranging this equation to solve for the radius (r) yields:

r = (mv) / (qB)

Step 3: Calculate for Each Isotope

Now, we can calculate the radius for both isotopes of iron (Fe⁵⁷ and Fe⁵⁸). The mass of each isotope can be calculated using the given mass number and the mass of a nucleon (1.6 x 10^-27 kg).

• For Fe⁵⁷: m = 57 * (1.6 x 10^-27 kg) = 9.12 x 10^-26 kg
• For Fe⁵⁸: m = 58 * (1.6 x 10^-27 kg) = 9.28 x 10^-26 kg

Next, we need to calculate the velocity (v) for both isotopes using the earlier derived formula:

v = sqrt((2 * 8.0 x 10^-17 J) / m)

Now, substituting the mass for each isotope:

• For Fe⁵⁷:

  v = sqrt((2 * 8.0 x 10^-17 J) / (9.12 x 10^-26 kg)) ≈ 4.18 x 10⁶ m/s

• For Fe⁵⁸:

  v = sqrt((2 * 8.0 x 10^-17 J) / (9.28 x 10^-26 kg)) ≈ 4.13 x 10⁶ m/s

Finally, we can calculate the radius for each isotope:

• For Fe⁵⁷:

  r = (9.12 x 10^-26 kg * 4.18 x 10⁶ m/s) / (1.6 x 10^-19 C * 20 x 10^-3 T) ≈ 0.055 m

• For Fe⁵⁸:

  r = (9.28 x 10^-26 kg * 4.13 x 10⁶ m/s) / (1.6 x 10^-19 C * 20 x 10^-3 T) ≈ 0.054 m

Summary of Results

The radii of the circular paths followed by the isotopes in the magnetic field are approximately:

• Fe⁵⁷: 0.055 m
• Fe⁵⁸: 0.054 m

This demonstrates how the mass of the isotopes affects their motion in a magnetic field, with the heavier isotope (Fe⁵⁸) having a slightly smaller radius due to its greater mass. If you have any further questions or need clarification on any of these steps, feel