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A lamp of power $942 \mathrm{~W}$ radiates energy uniformly in all direction. The wavelength of radiation is $660 \mathrm{~nm}$. The photon flux on a small screen $5.0 \mathrm{~m}$ from the lamp in units of $\frac{\text { photon }}{m^2 \cdot s}$ is
[Take Planck's constant $h=6.6 \times 10^{-34}$ SI unit)
Options:
[Take Planck's constant $h=6.6 \times 10^{-34}$ SI unit)
Solution:
2841 Upvotes
Verified Answer
The correct answer is:
$1 \times 10^{19}$
$\left.\mathrm{I}\right|_{\mathrm{r}=5 \mathrm{~m}}=\frac{\mathrm{P}}{\mathrm{A}}$
$$
=\frac{942}{\pi \times 5^2}=12
$$
So, no. of photon falling per sec
$$
\begin{aligned}
& =\frac{\mathrm{I}}{\frac{\mathrm{hc}}{\lambda}}=\frac{\mathrm{I} \lambda}{\mathrm{hc}} \\
& =\frac{12 \times 660 \times 10^{-9}}{6.6 \times 10^{-34} \times 3 \times 10^8} \\
& =4 \times 10^{19}
\end{aligned}
$$
So most suitable option is (d).
$$
=\frac{942}{\pi \times 5^2}=12
$$
So, no. of photon falling per sec
$$
\begin{aligned}
& =\frac{\mathrm{I}}{\frac{\mathrm{hc}}{\lambda}}=\frac{\mathrm{I} \lambda}{\mathrm{hc}} \\
& =\frac{12 \times 660 \times 10^{-9}}{6.6 \times 10^{-34} \times 3 \times 10^8} \\
& =4 \times 10^{19}
\end{aligned}
$$
So most suitable option is (d).
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