Question

A special kind of light bulb emits monochromatic light of 700nm. The electrical energy supply is at the rate of 60W and the bulb is 50 percent efficient that energy to light energy. How many photons are emitted by the bulb during its lifetime of one day

Answer 1

Hint: It is given to us that the bulb converts only fifty percent of the energy supplied to it into light energy. First, we will determine the energy of one photon using the expression of the energy of the photon. Further, we will determine the total energy emitted by the bulb if the form of photons and accordingly take the ratio to determine the number of photons emitted for a period of 1 day.

Formula used:
${{E}_{P}}=\dfrac{hc}{\lambda }$
$\eta =\dfrac{\text{Output power}}{\text{Input power}}$

Complete step-by-step solution:
It is given in the question that the efficiency($\eta $) of the bulb is 50 percent i.e. $\eta =0.5$.Efficiency in general is defined as the ratio of input power to that of the output power. This is mathematically written as,
$\eta =\dfrac{\text{Output power}}{\text{Input power}}$
The input power i.e. the rate of energy supply is given as 60W. Therefore the output power of the bulb i.e. the energy emitted by the bulb in the form of light per second is,
$\begin{align}
  & \eta =\dfrac{\text{Output power}}{\text{Input power}} \\
 & \Rightarrow 0.5=\dfrac{\text{Output power}}{60} \\
 & \Rightarrow \text{Output power}=0.5\times 60=30J/s \\
\end{align}$
The wavelength of light is given as 700nm. Let us say that the wavelength of a photon is denoted ($\lambda $ ). Then the energy (E) of the photon is given by,
$E=\dfrac{hc}{\lambda }$ where h is the Planck's constant i.e. $h=6.626\times {{10}^{-34}}J$ and c is the speed of light i.e. equal to $3\times {{10}^{8}}m/s$. hence from the above equation the energy of photon is given by,
$\begin{align}
  & E=\dfrac{hc}{\lambda } \\
 & \Rightarrow E=\dfrac{6.626\times {{10}^{-34}}Js\times 3\times {{10}^{8}}}{700nm} \\
 & \Rightarrow E=\dfrac{6.626\times {{10}^{-34}}Js\times 3\times {{10}^{8}}m/s}{700\times {{10}^{-9}}m} \\
 & \Rightarrow E=\dfrac{6.626\times {{10}^{-34}}\times 3\times {{10}^{15}}}{7}J \\
 & \Rightarrow E=\dfrac{6.626\times {{10}^{-34}}\times 3\times {{10}^{15}}}{7}J \\
 & \Rightarrow E=2.389\times {{10}^{-19}}J=2.4\times {{10}^{-19}}J \\
\end{align}$
The above energy obtained is the energy of one photon. The emitted bulb energy in the form of photons is equal to 30J/s. Therefore the number of photons emitted per second is,
$\begin{align}
  & \text{Number of photons/s}=\dfrac{30J/s}{2.4\times {{10}^{-19}}J}=12.5\times {{10}^{19}} \\
 & \Rightarrow \text{Number of photons/s}=125\times {{10}^{18}} \\
\end{align}$
Hence the number of photons emitted in a day is equal to,
$\begin{align}
  & \text{No}\text{. of photons/day}=\text{No}\text{. of photons/s}\times \text{No}\text{. of sec/day} \\
 & \Rightarrow \text{No}\text{. of photons/day}=125\times {{10}^{18}}\times 24\times 3600 \\
 & \Rightarrow \text{No}\text{. of photons/day}=108\times {{10}^{23}} \\
\end{align}$
Therefore the number of photons emitted by the bulb of wavelength 700nm is equal to $108\times {{10}^{23}}$.

Note: It is to be noted that energy emitted by the bulb is constant, hence the above equations are valid. Let us say the photon has a frequency of( $\gamma $). Therefore the energy of the photon can also be expressed as $E=h\gamma $.