PYQ NEET- Structure Of Atom L-6

Question: 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 $\mathrm{c}=3.0 \times 10^8 \mathrm{~ms}^{-1}$ ]#

A) $21.92 \mathrm{~cm}$

B) $219.3 \mathrm{~m}$

C) $219.2 \mathrm{~m}$

D) $2192 \mathrm{~m}$

Answer: $219.3 \mathrm{~m}$#

Solution:#

Energy of electromagnetic radiation (E) $$ =\frac{h c}{\lambda}=h \gamma $$

So, $\frac{c}{\lambda}=\gamma$ $$ \begin{aligned} & \Rightarrow \lambda=\frac{c}{\gamma} \ & \lambda=\frac{3 \times 10^8}{1368 \times 10^3}=219.3 \mathrm{~m} \end{aligned} $$