TS EAMCET · Physics · Dual Nature of Matter
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)
- A \(5 \times 10^{20}\)
- B \(2 \pi \times 10^{19}\)
- C \(\frac{6}{\pi} \times 10^{18}\)
- D \(1 \times 10^{19}\)
Answer & Solution
Correct Answer
(D) \(1 \times 10^{19}\)
Step-by-step Solution
Detailed explanation
\(\left.\mathrm{I}\right|_{\mathrm{r}=5 \mathrm{~m}}=\frac{\mathrm{P}}{\mathrm{A}}\) \[ =\frac{942}{\pi \times 5^2}=12 \] So, no. of photon falling per sec…
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