Question

A bulb of $40\text{W}$ is producing a light of wavelength $620\text{nm}$ with $80\%$ efficiency. Find the number of photons emitted by the bulb in $20$ seconds.
$(1\text{eV}=1.6\times {10}^{-19}\text{J},hc=12423\text{eV}\stackrel{o}{\text{A}})$

• A. $2\times {10}^{19}$
• B. $4\times {10}^{18}$
• C. $5\times {10}^{21}$
• D. $2\times {10}^{21}$

Answer 1

The correct option is D $2\times {10}^{21}$

We know, $\text{Energy}\left(E\right)=\text{Power (P)}\times \text{Time (t)}$
Efficiency of the Bulb = 80 %
As per the planck's quantum theory
E = $\frac{nhc}{\lambda }$ where $\lambda$ = wavelength of the light emitted by bulb and n is the number of photons
wavelength = 620 nm = $6200\stackrel{o}{\text{A}}$
Therefore,
$P\times t\times \text{efficiency}=n\times E$
$\Rightarrow 40\times 20\times \frac{80}{100}=n\times \frac{12423}{6200}\times 1.6\times {10}^{-19}$

$\Rightarrow n=2\times {10}^{21}$