Question

Estimate how many visible light photons a $120\text{W}$ light bulb emits per second. Assume the bulb has an efficiency of about $5\%$.
$(\lambda =500\text{nm},h=6.63\times {10}^{-34}\text{Joule-sec})$

• A. $1.508\times {10}^{19}$
• B. $15.08\times {10}^{19}$
• C. $0.15\times {10}^{19}$
• D. $150.8\times {10}^{19}$

Answer 1

The correct option is A $1.508\times {10}^{19}$

Energy emitted per second is,
$E=5\text{\%}\times 120\text{J}=6\text{J}$

If $N$ is the total number of photons emitted per second,

$E=h\nu \times N=\frac{hc}{\lambda }\times N$

$\Rightarrow N=\frac{E\lambda }{hc}$

$\Rightarrow N=\frac{6\times 500\times {10}^{-9}}{6.63\times {10}^{-34}\times 3.0\times {10}^8}$

$\Rightarrow N=1.508\times {10}^{19}$

Hence, option $\left(A\right)$ is correct.