
A 200W sodium street lamp emits yellow light of wavelength 0.6μm. Assuming it to be 25% efficient converting electrical energy to light, the number of photons of yellow light it emits per second is:
A.) $62\times {{10}^{20}}$
B.) $3\times {{10}^{19}}$
C.) $1.5\times {{10}^{20}}$
D.) $6\times {{10}^{18}}$
Answer
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Hint: Energy of a photon is inversely proportional to its wavelength. Ratio of total energy of all the photons to the power rating should give you the number of photons.
Formula used:
Energy of a single photon is given by,
$E=\dfrac{hc}{\lambda }$
Where,
h is the Planck constant,
c is the speed of light,
and $\lambda $ is the photon’s wavelength.
Complete step by step answer:
The power rating of the sodium lamp is 200W.
In other terms, the sodium lamp emits 200 Joules per second.
If, only 25% of the electrical energy is converted into light, the energy of the light emitted per second is given by,
${{E}_{1}}=200\times 0.25=50$Joules
As per the equation given above, energy of each proton is given by,
$E=\dfrac{hc}{\lambda }$............. Equation (1)
Where, $h=6.626\times {{10}^{-34}}{{m}^{2}}\dfrac{kg}{s}$,
$c=3\times {{10}^{8}}\dfrac{m}{s}$,
$\lambda =0.6\mu m=0.6\times {{10}^{-6}}m$
Putting these values in equation (1) we get,
$E=\dfrac{(6.626\times {{10}^{-34}})(3\times {{10}^{8}})}{(0.6\times {{10}^{-6}})}$,
Or, $E=3.313\times {{10}^{-19}}$ Joules
So, the number of photons per second is the ratio of total photon energy per second and energy of each photon.
Hence, the number of photons per second is given by,
$n=\dfrac{50}{3.313\times {{10}^{-19}}}$,
Or, $n=1.5\times {{10}^{20}}$
Hence, the correct answer option is (C).
Additional Information:
The question can also be solved using the frequency of the photon. In that case, question (1) should be modified in the following way,
$E=h\nu $
Note: The most common mistake in this kind of problem is the error in getting the correct dimensions. So, you should make sure that you have brought each constant to a specific dimension. For example, all the constants were brought into the SI units.
Formula used:
Energy of a single photon is given by,
$E=\dfrac{hc}{\lambda }$
Where,
h is the Planck constant,
c is the speed of light,
and $\lambda $ is the photon’s wavelength.
Complete step by step answer:
The power rating of the sodium lamp is 200W.
In other terms, the sodium lamp emits 200 Joules per second.
If, only 25% of the electrical energy is converted into light, the energy of the light emitted per second is given by,
${{E}_{1}}=200\times 0.25=50$Joules
As per the equation given above, energy of each proton is given by,
$E=\dfrac{hc}{\lambda }$............. Equation (1)
Where, $h=6.626\times {{10}^{-34}}{{m}^{2}}\dfrac{kg}{s}$,
$c=3\times {{10}^{8}}\dfrac{m}{s}$,
$\lambda =0.6\mu m=0.6\times {{10}^{-6}}m$
Putting these values in equation (1) we get,
$E=\dfrac{(6.626\times {{10}^{-34}})(3\times {{10}^{8}})}{(0.6\times {{10}^{-6}})}$,
Or, $E=3.313\times {{10}^{-19}}$ Joules
So, the number of photons per second is the ratio of total photon energy per second and energy of each photon.
Hence, the number of photons per second is given by,
$n=\dfrac{50}{3.313\times {{10}^{-19}}}$,
Or, $n=1.5\times {{10}^{20}}$
Hence, the correct answer option is (C).
Additional Information:
The question can also be solved using the frequency of the photon. In that case, question (1) should be modified in the following way,
$E=h\nu $
Note: The most common mistake in this kind of problem is the error in getting the correct dimensions. So, you should make sure that you have brought each constant to a specific dimension. For example, all the constants were brought into the SI units.
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