Question

A charged particle oscillates about its mean equilibrium position with a frequency of ${10}^{9}Hz.$ The electromagnetic waves produced:

Answer 1

As we know, a charged particle oscillates about its mean equilibrium position with a frequency of ${10}^{9}Hz.$

Therefore, the frequency of an electromagnetic wave produced by the oscillator is the same as that of a charged particle oscillating about its mean position i.e., ${10}^{9}Hz.$

The wavelength of light is given by,

$\lambda =\frac{c}{v}$ $...\left(i\right)$

Where,

$v={10}^{9}Hz$, frequency of electromagnetic wave

$c=3\times {10}^{8}m/s$

By putting the values in equation $\left(i\right)$ we get,

$\lambda =\frac{3\times {10}^8}{{10}^9}=0.3m$

As the frequency of electromagnetic wave is ${10}^{9}Hz,$ thus it will fall in the region of radio waves.

Final Answer: Option (a), (c) ,(d)