Question 147 of 165

Choose the correct option relating wavelengths of different parts of electromagnetic wave spectrum:

A${\lambda }_{\text{visible}}\lt {\lambda }_{\text{micro} \text{waves}}\lt {\lambda }_{\text{radio} \text{waves}}\lt {\lambda }_{\text{x-rays}}$B${\lambda }_{\text{radio} \text{waves}}\gt {\lambda }_{\text{micro} \text{waves}}\gt {\lambda }_{\text{visible}}\gt {\lambda }_{\text{x-rays}}$C${\lambda }_{\text{x-rays}}\lt {\lambda }_{\text{micro} \text{waves}}\lt {\lambda }_{\text{radio} \text{waves}}\lt {\lambda }_{\text{visible}}$D${\lambda }_{\text{visible}}\gt {\lambda }_{\text{x-rays}}\gt {\lambda }_{\text{radio} \text{waves}}\gt {\lambda }_{\text{micro} \text{waves}}$

Show answer and solution▾

Answer: (B) ${\lambda }_{\text{radio} \text{waves}}\gt {\lambda }_{\text{micro} \text{waves}}\gt {\lambda }_{\text{visible}}\gt {\lambda }_{\text{x-rays}}$

First, recall the spectrum of electromagnetic waves, and we know that in the case of electromagnetic waves the frequency is inversely proportional to wavelength.

Now from the above spectrum, we can say that, the relations of wavelength,
${\lambda }_{\text{radio waves}}\gt {\lambda }_{\text{micro waves}}\gt {\lambda }_{\text{visible}}\gt {\lambda }_{\text{x-rays}}$.

JEE Main Physics practice question on Electromagnetic Waves (Electromagnetic Spectrum), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.