Question

The Vividh Bharati station of All India. Radio, Delhi, broadcasts on a frequency of 1.368 kHz (kilo hertz). Calculate the wavelength of the electromagnetic
adiation emitted by transmitter. Which part of the electromagnetic spectrum does it belong to?​

Answer 1

Question :-

he Vividh Bharati station of All India. Radio, Delhi, broadcasts on a frequency of 1,368 kHz (kilo hertz). Calculate the wavelength of the electromagnetic adiation emitted by transmitter. Which part of the electromagnetic spectrum does it belong to?

AnSwer :-

frequency, v = 1368 kHz

speed of electromagnetic radiation,

c = 3 × 10⁸m/s

The wavelength, $\lambda$, is equal to c/v, where c is the speed of electromagnetic radiation in vaccum and v is the frequency.

substituting the given value , we have

$\sf \lambda = \dfrac{c}{v}$

$\sf = \dfrac{3.00 \times {10}^{8} {ms}^{ - 1} }{1368 \times {10}^{3} {s}^{ - 1} }$

$\sf = 219.3m$

This is a Characteristic radiowave wavelength

Answer 2

$\bigstar$ Given

The Vividh Bharati station of All India. Radio, Delhi, broadcasts on a frequency of 1,368 kHz (kilo hertz)

$\bigstar$ To Find

The wavelength of the electromagnetic radiation emitted by transmitter

Which part of the electromagnetic spectrum does it belong to

$\bigstar$ Solution

We know that

$\red{\sf \longrightarrow \lambda=\dfrac{c}{v}}$

Here

• λ = wavelength
• c = speed of light in vacuum
• v = frequency

Units

• λ = metre (m)
• c = m/s
• v = Hz

According to the question

We are asked to find the wavelength of the electromagnetic radiation emitted by transmitter

Therefore

We must find "λ"

Given that

The Vividh Bharati station of All India. Radio, Delhi, broadcasts on a frequency of 1,368 kHz (kilo hertz)

We know that

kilo (k) = 10³

Hence

• v = 1368 kHz = 1368 x 10³ Hz

We know that

• c = 3 x 10⁸ m/s

Substituting the values

We get

$\blue{\sf \longrightarrow \lambda=\dfrac{3 \times 10^{8}}{1368 \times 10^{3}} \ m}$

On further simplification

We get

$\green{\sf \longrightarrow \lambda=0.0021929 \times 10^{5} \ m}$

Hence

$\orange{\sf \longrightarrow \lambda=219.29 \ m}$

On approximation

We get

$\pink{\sf \longrightarrow \lambda \approx 219.3 \ m}$

Hence

$\star$ Wavelength = 219.3 m

Therefore

It belongs to "Radio waves region" of the electromagnetic spectrum