A particular station of All India Radio, New Delhi, broadcasts on a frequency of 1,368 kHz (kilohertz). The wavelength of the electromagnetic radiation emitted by the transmitter is :[speed of light, c=3.0×108 ms−1]

The correct option is 219.3 m.

To find the wavelength of the electromagnetic radiation emitted by the transmitter, we can use the wave relation that connects the speed of light, frequency, and wavelength:
$c=\nu \lambda$
where:
• $c$ is the speed of light in vacuum ($3.0\times {10}^8{\text{m s}}^{-1}$),
• $\nu$ is the frequency of the electromagnetic radiation,
• $\lambda$ is the wavelength.

Rearranging the formula to solve for the wavelength ($\lambda$), we get:
$\lambda =\frac{c}{\nu }$

First, let's convert the given frequency from kilohertz (kHz) to the SI unit, hertz (Hz or s^-1):
$\nu =1,368\text{kHz}=1,368\times {10}^3{\text{s}}^{-1}=1.368\times {10}^6{\text{s}}^{-1}$

Now, substitute the values of $c$ and $\nu$ into the rearranged formula:
$\lambda =\frac{3.0\times {10}^8{\text{m s}}^{-1}}{1.368\times {10}^6{\text{s}}^{-1}}$

Simplify the expression by dividing the coefficients and subtracting the exponents:
$\lambda =\left(\frac{3.0}{1.368}\right)\times {10}^{8-6}\text{m}$
$\lambda \approx 2.19298\times {10}^2\text{m}$
$\lambda \approx 219.3\text{m}$

Thus, the wavelength of the electromagnetic radiation emitted by the transmitter is approximately 219.3 m.