Question

Number of photons of wavelength 600 nm, emitted per second by an electric bulb of power 60 W is $(Take\,h=6\,\times\,10^{-36}Js)$

• A. 60
• B. 600
• C. $2\,\times\,10^{20}$
• D. $10^{20}$

Answer 1

Answer: (C)

Given that, the wavelength of photons
$ \lambda = 660\, nm = 660 \times 10^{ - 9 } \, m $
The energy of photons.
$E = n \,h\, v$
But $ E = P \times t $
$\therefore P \times t = n \, h \, v $
$ \Rightarrow P \times t = \frac{ n h c }{ \lambda } $
$\therefore 60 \times 1 = \frac{ n \times 6.6 \times 10^{ - 34} \times 3 \times 10^8 }{ 660 \times 10^{ - 9 }} $
$ \Rightarrow 60 = \frac{ n \times 6.6 \times 3 \times 10^{ - 19 }}{ 6.6} $
$ \Rightarrow n = \frac{ 60 }{ 3 \times 10^{ - 19 }} $
$ \Rightarrow \,= 2 \times 10^{ 20} $ photons