An AIR station is broadcasting the waves of wavelength 300 metres. If the radiating power of the transmitter is 10 kW, then the number of photons radiated per second is

$1.5 \times 10^{29}$

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$1.5 \times 10^{31}$

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$1.5 \times 10^{33}$

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$1.5 \times 10^{35}$

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Solution

The correct option is B
$1.5 \times 10^{31}$

$p = \frac{W}{t} = \frac{n h c}{λ t} \Rightarrow (\frac{n}{t}) = \frac{p λ}{h c} = \frac{10 \times 10^{3} \times 300}{6.6 \times 10^{- 34} \times 3 \times 10^{8}}$
$= 1.5 \times 10^{31}$

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An AIR station is broadcasting the waves of wavelength 300 metres. If the radiating power of the transmitter is 10 kW, then the number of photons radiated per second is

The correct option is B 1.5×1031 p=Wt=nhcλt⇒(nt)=pλhc=10×103×3006.6×10−34×3×108 =1.5×1031

The correct option is B
$1.5 \times 10^{31}$

$p = \frac{W}{t} = \frac{n h c}{λ t} \Rightarrow (\frac{n}{t}) = \frac{p λ}{h c} = \frac{10 \times 10^{3} \times 300}{6.6 \times 10^{- 34} \times 3 \times 10^{8}}$
$= 1.5 \times 10^{31}$