Question

If the frequency of an electromagnetic wave is 2 MHz, then the time period of oscillation of the accelerated charge is

• A. \( 2.5 \times 10^{-7} \) s
• B. \( 1 \times 10^{-7} \) s
• C. \( 5 \times 10^{-7} \) s
• D. \( 6 \times 10^{-7} \) s
• E. \( 2 \times 10^{-7} \) s

Hint:

The time period \( T \) is the inverse of the frequency. For high frequencies, the time period is very small.

Answer 1

The Correct Option is C

The time period \( T \) of an oscillating charge (or an electromagnetic wave) is related to the frequency \( f \) by the formula: \[ T = \frac{1}{f} \] Where: - \( T \) is the time period - \( f \) is the frequency We are given the frequency \( f = 2 \, \text{MHz} = 2 \times 10^6 \, \text{Hz} \). Now, calculate the time period \( T \): \[ T = \frac{1}{f} = \frac{1}{2 \times 10^6} = 5 \times 10^{-7} \, \text{s} \] Thus, the time period of oscillation is \( 5 \times 10^{-7} \) seconds. Hence, the correct answer is (C) \( 5 \times 10^{-7} \) s.