Question

The magnitude of an electric field which can just suspend a deuteron of mass \( 3.2 \times 10^{-27} \) kg freely in air is:

• A. \( 19.6 \times 10^{-8} \) NC\(^{-1}\)
• B. \( 196 \) NC\(^{-1}\)
• C. \( 1.96 \times 10^{-10} \) NC\(^{-1}\)
• D. \( 0.196 \) NC\(^{-1}\)

Hint:

To balance a charged particle in an electric field, use \( qE = mg \) where \( q \) is charge, \( m \) is mass, and \( g \) is gravitational acceleration.

Answer 1

The Correct Option is A

Step 1: Equilibrium Condition
A deuteron (a nucleus of deuterium) experiences two forces:
1. Gravitational force: \( F_g = mg \), where \( m = 3.2 \times 10^{-27} \) kg and \( g = 9.8 \) m/s\(^2\).
2. Electric force: \( F_e = qE \), where \( q \) is the charge of the deuteron.
For the deuteron to be suspended freely, these forces must be equal: \[ qE = mg \] Step 2: Substituting Values
The charge of a deuteron is equal to the charge of a proton:
\[ q = 1.6 \times 10^{-19} C \] The gravitational force is: \[ F_g = (3.2 \times 10^{-27} \text{ kg}) \times (9.8 \text{ m/s}^2) \] \[ F_g = 3.136 \times 10^{-26} \text{ N} \] Step 3: Solving for Electric Field \[ E = \frac{mg}{q} = \frac{3.136 \times 10^{-26}}{1.6 \times 10^{-19}} \] \[ E = 19.6 \times 10^{-8} \text{ NC}^{-1} \] Step 4: Conclusion
Thus, the required electric field to suspend the deuteron is \( 19.6 \times 10^{-8} \) NC\(^{-1}\).