Singly ionized (one electron removed) atoms are accelerated and then passed through a velocity selector consisting of perpendicular electric and magnetic fields. The electric field is 155 V/m and the magnetic field is 0.0315 T. The ions next enter a uniform magnetic field of magnitude 0.0175 T that is oriented perpendicular to their velocity. (a) How fast are the ions moving when they emerge from the velocity selector? (b) If the radius of the path of the ions in the second magnetic field is 17.5 cm, what is their mass?

Verified step by step guidance

1. To find the speed of the ions as they emerge from the velocity selector, we use the condition for the velocity selector where the electric force equals the magnetic force. The formula for this condition is \( v = \frac{E}{B} \), where \( E \) is the electric field and \( B \) is the magnetic field.
2. Substitute the given values into the formula: \( E = 155 \text{ V/m} \) and \( B = 0.0315 \text{ T} \). Calculate \( v \) using \( v = \frac{155}{0.0315} \). This will give you the speed of the ions as they emerge from the velocity selector.
3. For part (b), we need to find the mass of the ions using the radius of their path in the second magnetic field. The formula for the radius \( r \) of the path of a charged particle in a magnetic field is \( r = \frac{mv}{qB} \), where \( m \) is the mass, \( v \) is the velocity, \( q \) is the charge, and \( B \) is the magnetic field.
4. Rearrange the formula to solve for mass \( m \): \( m = \frac{rqB}{v} \). Substitute the known values: \( r = 0.175 \text{ m} \), \( B = 0.0175 \text{ T} \), \( v \) from part (a), and \( q \) is the charge of a singly ionized atom, which is the elementary charge \( 1.6 \times 10^{-19} \text{ C} \).
5. Calculate \( m \) using the rearranged formula \( m = \frac{0.175 \times 0.0175 \times 1.6 \times 10^{-19}}{v} \). This will give you the mass of the ions.

Key Concepts

Velocity Selector

A velocity selector uses perpendicular electric and magnetic fields to filter particles based on their velocity. The electric field exerts a force on charged particles, while the magnetic field exerts a perpendicular force. Only particles with a specific velocity, where the electric and magnetic forces balance, pass through without deflection. This velocity is given by v = E/B, where E is the electric field strength and B is the magnetic field strength.

Lorentz Force

The Lorentz force is the force experienced by a charged particle moving through electric and magnetic fields. It is the sum of the electric force (F = qE) and the magnetic force (F = qvB), where q is the charge, v is the velocity, E is the electric field, and B is the magnetic field. In a velocity selector, these forces are set to balance each other, allowing particles with a specific velocity to pass through.

Circular Motion in Magnetic Fields

When charged particles enter a uniform magnetic field perpendicular to their velocity, they experience a centripetal force that causes them to move in a circular path. The radius of this path is determined by the equation r = mv/qB, where m is the mass, v is the velocity, q is the charge, and B is the magnetic field strength. This relationship allows for the determination of particle mass when other variables are known.