Question

A 100 watt bulb emits monochromatic light of wavelength 400 nm. Calculate

the number of photons emitted per second by the bulb.

Answer 1

Given,
Power of the bulb = 100 watt = work/time = 100 J/s
$$\begin{gathered} Energyofoneproton, \\ E=h\frac{c}{\lambda } \\ h=6.626\times {10}^{-34}Js \\ c=3\times {10}^{8}m/s \\ \lambda =400\times {10}^{-9}m \\ Therefore,E=\frac{6.626\times {10}^{-34}Js\times 3\times {10}^{8}m/s}{400\times {10}^{-9}m} \\ =4.969\times {10}^{-19}J \\ No.ofphotosemittedpersecond \\ =\frac{100J/s}{4.969\times {10}^{-19}J}=\mathbf{2}\mathbf{.}\mathbf{012}\mathbf{\times }{\mathbf{10}}^{\mathbf{20}}\mathbf{}{\mathit{s}}^{\mathbf{-}\mathbf{1}} \end{gathered}$$

12.
$$\begin{gathered} \mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }} \\ \mathrm{or},\mathrm{\lambda }=\frac{\mathrm{hc}}{\mathrm{E}} \\ \mathrm{E}=1\mathrm{eV}=1.60\times {10}^{-19}\mathrm{J} \\ \mathrm{Therefore},\mathrm{E}=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{1.60\times {10}^{-19}}=12.42\times {10}^{-7}\mathrm{m}=0.0124\times {10}^{-10}\mathrm{m}=0.0124\stackrel{°}{\mathrm{A}} \end{gathered}$$