Question

If number of photons emitted by a bulb of $40\, W$ in $1\, \min$ with $50 \%$ efficiency will be approximately $X \times 10^{21}$, then the value of $X$ will be $(\lambda=6200\,\mathring{A}, h c=12400\, eV\, \mathring{A})$

Answer 1

$\Delta E=\frac{12400}{6200}=2\, eV$
$2 \times 1.6 \times 10^{-19} \times n=40 \times 60 \times 0.5$
$n=3.75 \times 10^{21}$