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A 150 watt bulb emits light of wavelength $6600 Å$ and only $8 \%$ of the energy is emitted as light. How many photons are emitted by the bulb per second?
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Verified Answer
The correct answer is:
$4 \times 10^{19}$
Power of the bulb $=150 \mathrm{~W}=150 \mathrm{~J} \mathrm{~s}^{-1}$
As only $8 \%$ of the energy is emitted as light so, the total energy emitted per second
$$
=\frac{150 \mathrm{~J} \times 8}{100}=12 \mathrm{~J}
$$
Energy of one photon, $E=h v=\frac{h c}{\lambda}$
$$
\begin{aligned}
&=\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right) \times\left(3 \times 10^{8} \mathrm{~ms}^{-1}\right)}{6600 \times 10^{-10} \mathrm{~m}} \\
&=3.0118 \times 10^{-19} \mathrm{~J}
\end{aligned}
$$
$\therefore$ Number of photons emitted per second
$$
=\frac{12 \mathrm{~J}}{3.0118 \times 10^{-19} \mathrm{~J}}=3.98 \times 10^{19} \approx 4.0 \times 10^{19}
$$
As only $8 \%$ of the energy is emitted as light so, the total energy emitted per second
$$
=\frac{150 \mathrm{~J} \times 8}{100}=12 \mathrm{~J}
$$
Energy of one photon, $E=h v=\frac{h c}{\lambda}$
$$
\begin{aligned}
&=\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right) \times\left(3 \times 10^{8} \mathrm{~ms}^{-1}\right)}{6600 \times 10^{-10} \mathrm{~m}} \\
&=3.0118 \times 10^{-19} \mathrm{~J}
\end{aligned}
$$
$\therefore$ Number of photons emitted per second
$$
=\frac{12 \mathrm{~J}}{3.0118 \times 10^{-19} \mathrm{~J}}=3.98 \times 10^{19} \approx 4.0 \times 10^{19}
$$
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