A blue lamp mainly emits light of wavelength.\[4500\mathop {\,A}\limits^o \] The lamp is rated at \[150\,{\text{W}}\] and \[{\text{8}}\% \] of the energy is emitted as visible light. The number of photons emitted by the lamp per second is:
A. \[3 \times {10^{19}}\]
B. \[3 \times {10^{24}}\]
C. \[3 \times {10^{20}}\]
D. \[3 \times {10^{18}}\]
Answer
652.2k+ views
Hint:Calculate the energy of the total photons in the emitted visible light. Derive the relation between the power, energy, wavelength and number of photons.
Formulae used:
The energy \[E\] of a single photon is given by
\[E = \dfrac{{hc}}{\lambda }\] …… (1)
Here, \[h\] is the Planck’s constant, \[c\] is the speed of light and \[\lambda \] is the wavelength of the photon.
The relation between the power \[P\] and energy \[E\] is
\[P = \dfrac{E}{t}\] …… (2)
Here, \[t\] is the time.
Complete step by step answer:
Calculate the energy \[{E_n}\] of \[n\] number of photons in the emitted light.
\[{E_n} = nE\]
Substitute \[\dfrac{{hc}}{\lambda }\] for \[E\] in the above equation.
\[{E_n} = n\dfrac{{hc}}{\lambda }\] …… (3)
Calculate the total energy of the lamp.
Rewrite equation (2) for the total energy \[{E_T}\] of the lamp emitted in one second.
\[{E_T} = {P_T}t\]
Here, \[{P_T}\] is the total power of the lamp.
Substitute \[150\,{\text{W}}\] for \[{P_T}\] and \[1\,{\text{s}}\] for \[t\] in the above equation.
\[{E_T} = \left( {150\,{\text{W}}} \right)\left( {1\,{\text{s}}} \right)\]
\[ \Rightarrow {E_T} = 150\,{\text{J}}\]
Only \[{\text{8}}\% \] of the total energy of the lamp is emitted as the visible light.
Hence, the total energy \[{E_L}\] of the photons in the emitted light is \[{\text{8}}\% \] of the total energy \[{E_T}\] of the lamp.
\[{E_L} = \dfrac{8}{{100}}{E_T}\] …… (4)
The energy \[{E_n}\] of the \[n\] number of photons in the emitted light is equal to the total energy \[{E_L}\] of the photons in the emitted light.
\[{E_n} = {E_L}\]
Substitute \[n\dfrac{{hc}}{\lambda }\] for \[{E_n}\] and \[\dfrac{8}{{100}}{E_T}\] for \[{E_L}\] in the above equation.
\[n\dfrac{{hc}}{\lambda } = \dfrac{8}{{100}}{E_T}\]
Rearrange the above equation for \[n\].
\[n = \dfrac{8}{{100}}\dfrac{{{E_T}\lambda }}{{hc}}\]
Substitute \[150\,{\text{J}}\] for \[{E_T}\], \[4500\mathop {\,A}\limits^o \] for \[\lambda \], \[6.63 \times {10^{ - 34}}\,{\text{J}} \cdot {\text{s}}\] for \[h\] and \[3 \times {10^8}\,{\text{m/s}}\] for \[c\] in the above equation.
\[n = \dfrac{8}{{100}}\dfrac{{\left( {150\,{\text{J}}} \right)\left( {4500\mathop {\,A}\limits^o } \right)}}{{\left( {6.63 \times {{10}^{ - 34}}\,{\text{J}} \cdot {\text{s}}} \right)\left( {3 \times {{10}^8}\,{\text{m/s}}} \right)}}\]
\[ \Rightarrow n = \dfrac{8}{{100}}\dfrac{{\left( {150\,{\text{J}}} \right)\left( {4500 \times {{10}^{ - 10}}\,{\text{m}}} \right)}}{{\left( {6.63 \times {{10}^{ - 34}}\,{\text{J}} \cdot {\text{s}}} \right)\left( {3 \times {{10}^8}\,{\text{m/s}}} \right)}}\]
\[ \Rightarrow n = 2.71 \times {10^{19}}\]
\[ \Rightarrow n \approx 3 \times {10^{19}}\]
Therefore, the number of the photons emitted by the lamp per second is \[3 \times {10^{19}}\].
Hence, the correct option is A.
Note:While doing solution please make sure that you have calculate the total number of photons emitted by the lamp using the energy \[{\text{8}}\% \]of the light emitted by the lamp and not the total energy of the lamp because this the place where mistake could be happened.
Formulae used:
The energy \[E\] of a single photon is given by
\[E = \dfrac{{hc}}{\lambda }\] …… (1)
Here, \[h\] is the Planck’s constant, \[c\] is the speed of light and \[\lambda \] is the wavelength of the photon.
The relation between the power \[P\] and energy \[E\] is
\[P = \dfrac{E}{t}\] …… (2)
Here, \[t\] is the time.
Complete step by step answer:
Calculate the energy \[{E_n}\] of \[n\] number of photons in the emitted light.
\[{E_n} = nE\]
Substitute \[\dfrac{{hc}}{\lambda }\] for \[E\] in the above equation.
\[{E_n} = n\dfrac{{hc}}{\lambda }\] …… (3)
Calculate the total energy of the lamp.
Rewrite equation (2) for the total energy \[{E_T}\] of the lamp emitted in one second.
\[{E_T} = {P_T}t\]
Here, \[{P_T}\] is the total power of the lamp.
Substitute \[150\,{\text{W}}\] for \[{P_T}\] and \[1\,{\text{s}}\] for \[t\] in the above equation.
\[{E_T} = \left( {150\,{\text{W}}} \right)\left( {1\,{\text{s}}} \right)\]
\[ \Rightarrow {E_T} = 150\,{\text{J}}\]
Only \[{\text{8}}\% \] of the total energy of the lamp is emitted as the visible light.
Hence, the total energy \[{E_L}\] of the photons in the emitted light is \[{\text{8}}\% \] of the total energy \[{E_T}\] of the lamp.
\[{E_L} = \dfrac{8}{{100}}{E_T}\] …… (4)
The energy \[{E_n}\] of the \[n\] number of photons in the emitted light is equal to the total energy \[{E_L}\] of the photons in the emitted light.
\[{E_n} = {E_L}\]
Substitute \[n\dfrac{{hc}}{\lambda }\] for \[{E_n}\] and \[\dfrac{8}{{100}}{E_T}\] for \[{E_L}\] in the above equation.
\[n\dfrac{{hc}}{\lambda } = \dfrac{8}{{100}}{E_T}\]
Rearrange the above equation for \[n\].
\[n = \dfrac{8}{{100}}\dfrac{{{E_T}\lambda }}{{hc}}\]
Substitute \[150\,{\text{J}}\] for \[{E_T}\], \[4500\mathop {\,A}\limits^o \] for \[\lambda \], \[6.63 \times {10^{ - 34}}\,{\text{J}} \cdot {\text{s}}\] for \[h\] and \[3 \times {10^8}\,{\text{m/s}}\] for \[c\] in the above equation.
\[n = \dfrac{8}{{100}}\dfrac{{\left( {150\,{\text{J}}} \right)\left( {4500\mathop {\,A}\limits^o } \right)}}{{\left( {6.63 \times {{10}^{ - 34}}\,{\text{J}} \cdot {\text{s}}} \right)\left( {3 \times {{10}^8}\,{\text{m/s}}} \right)}}\]
\[ \Rightarrow n = \dfrac{8}{{100}}\dfrac{{\left( {150\,{\text{J}}} \right)\left( {4500 \times {{10}^{ - 10}}\,{\text{m}}} \right)}}{{\left( {6.63 \times {{10}^{ - 34}}\,{\text{J}} \cdot {\text{s}}} \right)\left( {3 \times {{10}^8}\,{\text{m/s}}} \right)}}\]
\[ \Rightarrow n = 2.71 \times {10^{19}}\]
\[ \Rightarrow n \approx 3 \times {10^{19}}\]
Therefore, the number of the photons emitted by the lamp per second is \[3 \times {10^{19}}\].
Hence, the correct option is A.
Note:While doing solution please make sure that you have calculate the total number of photons emitted by the lamp using the energy \[{\text{8}}\% \]of the light emitted by the lamp and not the total energy of the lamp because this the place where mistake could be happened.
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