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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?
Solution:
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Verified Answer
We know that, energy of photons $\mathrm{E}=\mathrm{nh} v$
Let $\mathrm{n}_{\mathrm{A}}$ is the number of photons falling per second of beam $\mathrm{A}$ and $\mathrm{n}_{\mathrm{B}}$ is the number of photons fallings per second of beam $\mathrm{B}$.
So, $\mathrm{n}_{\mathrm{A}}=2 \mathrm{n}_{\mathrm{B}}$
Energy of falling photon of beam $\mathrm{E}_{\mathrm{A}}=\mathrm{hv}_{\mathrm{A}} \mathrm{n}_{\mathrm{A}}$
Energy of falling photon of beam $E_B=h v_B n_B$
Now, according to question,
enrgy or intensity of $\mathrm{A}=$ energy or intensity of $\mathrm{B}$
$$
\begin{aligned}
&\therefore \mathrm{n}_{\mathrm{A}} \mathrm{hv} v_{\mathrm{A}}=\mathrm{n}_{\mathrm{B}} \mathrm{hv} v_{\mathrm{B}} \\
&\Rightarrow \quad \frac{v_{\mathrm{A}}}{v_{\mathrm{B}}}=\frac{\mathrm{n}_{\mathrm{B}}}{\mathrm{n}_{\mathrm{A}}}=\frac{\mathrm{n}_{\mathrm{B}}}{2 \mathrm{n}_{\mathrm{B}}}=\frac{1}{2} \\
&\Rightarrow \quad v_{\mathrm{B}}=2 v_{\mathrm{A}}
\end{aligned}
$$
So the frequency of source beam $B$ is twice the source of frequency beam $\mathrm{A}$.
Let $\mathrm{n}_{\mathrm{A}}$ is the number of photons falling per second of beam $\mathrm{A}$ and $\mathrm{n}_{\mathrm{B}}$ is the number of photons fallings per second of beam $\mathrm{B}$.
So, $\mathrm{n}_{\mathrm{A}}=2 \mathrm{n}_{\mathrm{B}}$
Energy of falling photon of beam $\mathrm{E}_{\mathrm{A}}=\mathrm{hv}_{\mathrm{A}} \mathrm{n}_{\mathrm{A}}$
Energy of falling photon of beam $E_B=h v_B n_B$
Now, according to question,
enrgy or intensity of $\mathrm{A}=$ energy or intensity of $\mathrm{B}$
$$
\begin{aligned}
&\therefore \mathrm{n}_{\mathrm{A}} \mathrm{hv} v_{\mathrm{A}}=\mathrm{n}_{\mathrm{B}} \mathrm{hv} v_{\mathrm{B}} \\
&\Rightarrow \quad \frac{v_{\mathrm{A}}}{v_{\mathrm{B}}}=\frac{\mathrm{n}_{\mathrm{B}}}{\mathrm{n}_{\mathrm{A}}}=\frac{\mathrm{n}_{\mathrm{B}}}{2 \mathrm{n}_{\mathrm{B}}}=\frac{1}{2} \\
&\Rightarrow \quad v_{\mathrm{B}}=2 v_{\mathrm{A}}
\end{aligned}
$$
So the frequency of source beam $B$ is twice the source of frequency beam $\mathrm{A}$.
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