Question

If $10\text{\%}$ of the electrical energy supplied to an incandescent light is radiated as visible light, the number of photons emitted by a $60\text{W}$ bulb per second is $1.8\times {10}^{9.5k}$. The value of $k$ is(Integer only) $(\lambda =6000\dot{A})$

Answer 1

Light Energy per second,

$E=\frac{10}{100}\times 60=6\text{J}=\frac{6}{1.6\times {10}^{-19}}\text{eV}$

$\therefore E=\text{Number of photons}\times \text{energy of each photon}$

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

$\Rightarrow \frac{6}{1.6\times {10}^{-19}}=N\times \frac{12400\text{eV}\dot{A}}{6000\dot{A}}$

$\Rightarrow 6=N\times \frac{12400}{6000}\times 1.6\times {10}^{-19}$

$\Rightarrow N=1.8\times {10}^{19}$ photons per second

Given, $N=1.8\times {10}^{9.5k}$

$\therefore k=2$