JEE MAIN - Physics (2022 - 30th June Morning Shift - No. 17)
A source of monochromatic light liberates 9 $$\times$$ 1020 photon per second with wavelength 600 nm when operated at 400 W. The number of photons emitted per second with wavelength of 800 nm by the source of monochromatic light operating at same power will be :
12 $$\times$$ 1020
6 $$\times$$ 1020
9 $$\times$$ 1020
24 $$\times$$ 1020
Wyjaśnienie
As we know
$$ \begin{aligned} &I=\frac{E}{A t}=\frac{n h v}{A t} \Rightarrow \frac{n}{t}=\frac{I A \lambda}{h C} \Rightarrow \frac{n}{t}=\frac{\rho \lambda}{h c} \Rightarrow \frac{n}{t}=\rho \lambda \\\\ &\Rightarrow\left(\frac{n}{t}\right)_{2}=\left(\frac{n}{t}\right)_{1} \times \frac{p_{2} \lambda_{2}}{p_{1} \lambda 1}=9 \times 10^{20} \times \frac{P}{P} \times \frac{800}{600}=12 \times 10^{20} \end{aligned} $$
$$ \begin{aligned} &I=\frac{E}{A t}=\frac{n h v}{A t} \Rightarrow \frac{n}{t}=\frac{I A \lambda}{h C} \Rightarrow \frac{n}{t}=\frac{\rho \lambda}{h c} \Rightarrow \frac{n}{t}=\rho \lambda \\\\ &\Rightarrow\left(\frac{n}{t}\right)_{2}=\left(\frac{n}{t}\right)_{1} \times \frac{p_{2} \lambda_{2}}{p_{1} \lambda 1}=9 \times 10^{20} \times \frac{P}{P} \times \frac{800}{600}=12 \times 10^{20} \end{aligned} $$