Ncert Exercises & Solutions

Two monochromatic beams A and B of equal intensity I, hit a screen. The number of photons hitting the screen by beam A is twice that by beam B. Then, what inference can you make about their frequencies?

Hint: The intensity of the radiation depends on the frequency of the radiation.

Step 1: Find the energy of photons of beam A and beam B.

Suppose n $_{A}$ is the number of photons falling per second of beam A and n $_{B}$ is the number of photons falling per second of beam B.

Thus, $n_{A} = 2 n_{B}$

The energy of falling photons of beam A $= {hν}_{A}$

The energy of falling photons of beam B=h $ν_{B}$

Step 2: Equate their intensities.

Now, according to the question,

The intensity of A= intensity of B

$\begin{aligned} n_{A} {hν}_{A} & = n_{B} {hν}_{B} \\ \Rightarrow \frac{ν_{A}}{ν_{B}} & = \frac{n_{B}}{n_{A}} = \frac{n_{B}}{2 n_{B}} = \frac{1}{2} \\ \Rightarrow ν_{B} & = 2 ν_{A} \end{aligned}$

Thus, from this relation, we can that frequency of beam B is twice that of beam A.

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions