Question

Resting potential of axonal membrane is restored mainly due to the --------------

• A. rapid diffusion of sodium ions to inside of the membrane.
• B. rapid diffusion of potassium ions to outside of the membrane.
• C. rapid diffusion of potassium ions to inside of the membrane.
• D. slow diffusion of potassium ions to inside of the membrane.

Hint:

After depolarisation, potassium ions rush out of the axon and make the inside negative again.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
During a nerve impulse, \(\text{Na}^{+}\) channels open and \(\text{Na}^{+}\) enters, reversing the polarity of the membrane.

Step 2: Repolarisation
Soon after, \(\text{Na}^{+}\) channels close and voltage-gated \(\text{K}^{+}\) channels open. \(\text{K}^{+}\) ions diffuse rapidly to the outside of the axon.

Step 3: Result
Loss of positive charge makes the inside of the membrane negative again, so the resting potential is restored. The \(\text{Na}^{+}\)-\(\text{K}^{+}\) pump later resets the ion concentrations.

Final Answer:
The resting potential is restored by rapid outward diffusion of potassium ions. \[ \boxed{\text{(B) }\text{rapid diffusion of potassium ions to outside of the membrane.}} \]