Question

A bulb of 40 W is producing a light of wavelength 620 nm with 80% of efficiency, then the number of photons emitted by the bulb in 20 seconds are ($1eV=1.6\times {10}^{-19}$ J, $hc=12400eV$) :

• A. $2\times {10}^{18}$
• B. ${10}^{18}$
• C. ${10}^{21}$
• D. $2\times {10}^{11}$

Answer 1

Answer: (D)

$2\times {10}^{11}$

Energy that will convert into the photons per second, $\frac{80\times 40}{100}=32J$.

Energy that will convert into the photons in 20 seconds, $E_T=\frac{80\times 40}{100}\times 20=640J$.

Now we need to calculate how much photons will produce in this energy of wavelength $\lambda =620$ nm. Let us say n photons will be produced.
Then,
$n=\frac{E_T}{\frac{hc}{\lambda }}=\frac{640}{\frac{12400\times 1.6\times {10}^{-19}}{620\times {10}^{-9}}}=2.0\times {10}^{11}$

Option 'D' is correct.