A blue lamp is rated at 150 W and 8% of the energy is emitted as blue light at λ = 4500 Å. The number of photons emitted by the lamp per second is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 2.71 × 1019
Only the part of the power emitted as blue light produces the photons counted here. The number of photons per second is this light power divided by the energy of one photon.
\( \displaystyle P_{\mathrm{light}} = 8\% \times 150\,\mathrm{W} = 12\,\mathrm{W} \)
\( \displaystyle E = \frac{hc}{\lambda} = \frac{6.63\times10^{-34}\,\mathrm{J\,s} \times 3\times10^{8}\,\mathrm{m\,s^{-1}}}{4500\times10^{-10}\,\mathrm{m}} \)
\( \displaystyle E \approx 4.42\times10^{-19}\,\mathrm{J} \)
\( \displaystyle N = \frac{P_{\mathrm{light}}}{E} = \frac{12\,\mathrm{J\,s^{-1}}}{4.42\times10^{-19}\,\mathrm{J}} \)
\( \displaystyle N \approx 2.71\times10^{19}\,\mathrm{s^{-1}} \)
Shortcut: With \( E(\mathrm{eV}) = \frac{12400}{\lambda(\text{in angstrom})} \), a photon of 4500 angstrom has about 2.76 eV \( = 4.42\times10^{-19} \) J, and \( 12/4.42\times10^{-19} \approx 2.71\times10^{19} \).