Problem 69 - A Complete Resource Book in Physics for JEE Main

Question

A body has a charge of one coulomb. The number of excess (or lesser) electrons on it from its normal state will be (A) \(\infty\) (B) \(1.6 \times 10^{-19}\) (C) \(1.6 \times 10^{19}\) (D) \(6.25 \times 10^{18}\)

Step-by-Step Solution

Answer

The number of excess (or lesser) electrons on the body from its normal state is \(6.25 × 10^{18}\). So, the correct answer is (D) \(6.25 \times 10^{18}\).

Step 1: Identifying the given information

We know the following information: - Charge of the body, Q = 1 C (coulomb) - Charge of one electron, e = \(-1.6 \times 10^{-19}\) C - We need to find the number of excess or lesser electrons, n Note that the charge of an electron is negative, indicating that the body has an excess of electrons.

Step 2: Use the formula of charge

The formula to calculate the charge of a body with a certain number of excess or lesser electrons is: Q = n × e Where, Q = charge of the body n = number of excess or lesser electrons e = charge of one electron Our given information is: Q = 1 C e = \(-1.6 \times 10^{-19}\) C We need to find n.

Step 3: Solve for n

We can now solve the equation Q = n × e for n: 1 C = n × \((-1.6 \times 10^{-19})\) C To find n, divide both sides by \(-1.6 \times 10^{-19}\) C: n = \(\frac{1 C}{-1.6 \times 10^{-19} C}\)

Step 4: Calculate the number of excess or lesser electrons

We can now calculate the value of n: n = \(\frac{1}{-1.6 \times 10^{-19}}\) n = \(6.25 × 10^{18}\) The number of excess (or lesser) electrons on the body from its normal state is \(6.25 × 10^{18}\). So, the correct answer is (D) \(6.25 \times 10^{18}\).