A 200 W sodium street lamp emits yellow light of wavelength 0.6 μm. Assuming it to be 25% efficient in 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
602.1k+ views
Hint
Here, we use the concept of Planck’s energy which states that the energy of a photon which is emitted due to transition of light is given by, $E = nh\upsilon = \dfrac{{nhc}}{\lambda }$, where n is the number of photons (an integer), h is the Planck’s constant ($h = 6.626 \times {10^{ - 34}}Js$), c is the speed of light ($c = 3.0 \times {10^8}m/s$), and λ is the wavelength of the light.
And the concept of ‘power’, which is defined as the rate of energy, i. e., $P = \dfrac{E}{t}$.
We also use the formula for efficiency (η) given by, $\eta = \dfrac{{{P_{out}}}}{{{P_{in}}}}$, where Pout is the output power and Pin is the input power.
Complete step by step answer
It is given that, the power of the sodium street lamp is, $P_{in}$ = 200 W;
Its efficiency, η = 25% = 0.25.
Calculate the output power ($P_{out}$) of the bulb, which is the power converted to light, by using the efficiency (η) formula given by, $\eta = \dfrac{{{P_{out}}}}{{{P_{in}}}}$.
$\Rightarrow {P_{out}} = \eta \times {P_{in}}$.
Put the given values.
$\Rightarrow {P_{out}} = 0.25 \times 200W\\\therefore {P_{out}} = 50W$
This implies that the energy of photons emitted by the bulb is 50 J/s.
Therefore, E = 50 J.
Now, consider the energy formula,
$\Rightarrow E = nh\upsilon = \dfrac{{nhc}}{\lambda }$.
Rearrange the equation in terms of n (number of photons).
$\Rightarrow n = \dfrac{{E\lambda }}{{hc}}$
Substitute the values.
$\Rightarrow n = \dfrac{{50J \times (0.6 \times {{10}^{ - 6}}m)}}{{(6.626 \times {{10}^{ - 34}}Js) \times (3.0 \times {{10}^8}m/s)}}\\ \Rightarrow n = \dfrac{{{{10}^{ - 6}}}}{{6.626 \times {{10}^{ - 34 + 8}}}}\\\therefore n = 1.5 \times {10^{20}}$
Thus, the number of photons of yellow light emitted by the bulb per second is $1.5 \times {10^{20}}$. The correct answer is option (C).
Note
Planck's postulate is one of the fundamental principles of quantum mechanics. It states that the energy of oscillators in a black body is quantized and is given by $E = nh\upsilon = \dfrac{{nhc}}{\lambda }$.
Here, we use the concept of Planck’s energy which states that the energy of a photon which is emitted due to transition of light is given by, $E = nh\upsilon = \dfrac{{nhc}}{\lambda }$, where n is the number of photons (an integer), h is the Planck’s constant ($h = 6.626 \times {10^{ - 34}}Js$), c is the speed of light ($c = 3.0 \times {10^8}m/s$), and λ is the wavelength of the light.
And the concept of ‘power’, which is defined as the rate of energy, i. e., $P = \dfrac{E}{t}$.
We also use the formula for efficiency (η) given by, $\eta = \dfrac{{{P_{out}}}}{{{P_{in}}}}$, where Pout is the output power and Pin is the input power.
Complete step by step answer
It is given that, the power of the sodium street lamp is, $P_{in}$ = 200 W;
Its efficiency, η = 25% = 0.25.
Calculate the output power ($P_{out}$) of the bulb, which is the power converted to light, by using the efficiency (η) formula given by, $\eta = \dfrac{{{P_{out}}}}{{{P_{in}}}}$.
$\Rightarrow {P_{out}} = \eta \times {P_{in}}$.
Put the given values.
$\Rightarrow {P_{out}} = 0.25 \times 200W\\\therefore {P_{out}} = 50W$
This implies that the energy of photons emitted by the bulb is 50 J/s.
Therefore, E = 50 J.
Now, consider the energy formula,
$\Rightarrow E = nh\upsilon = \dfrac{{nhc}}{\lambda }$.
Rearrange the equation in terms of n (number of photons).
$\Rightarrow n = \dfrac{{E\lambda }}{{hc}}$
Substitute the values.
$\Rightarrow n = \dfrac{{50J \times (0.6 \times {{10}^{ - 6}}m)}}{{(6.626 \times {{10}^{ - 34}}Js) \times (3.0 \times {{10}^8}m/s)}}\\ \Rightarrow n = \dfrac{{{{10}^{ - 6}}}}{{6.626 \times {{10}^{ - 34 + 8}}}}\\\therefore n = 1.5 \times {10^{20}}$
Thus, the number of photons of yellow light emitted by the bulb per second is $1.5 \times {10^{20}}$. The correct answer is option (C).
Note
Planck's postulate is one of the fundamental principles of quantum mechanics. It states that the energy of oscillators in a black body is quantized and is given by $E = nh\upsilon = \dfrac{{nhc}}{\lambda }$.
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