Question

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

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

Answer 1

Answer: (C)

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