
A modern \[200W\] sodium street lamp emits yellow light of wavelength $0.6\mu m$. Assuming it to be $25\%$ efficient in converting electrical energy into light, the number of photons of yellow light it emits per second is:
A) $62 \times {10^{29}}$
B) $3 \times {10^{19}}$
C) $1.5 \times {10^{20}}$
D) $6 \times {10^{18}}$
Answer
213.6k+ views
Hint: Recall the photoelectric effect of light. As per this effect, when light is incident on the surface of a metal, then the electrons are emitted. The ejected electrons are thus known as photo electrons. The electrons are ejected from the metal surface because they absorb energy.
Complete step by step solution:
Given that $P = 200W$
wavelength$\lambda = 0.6\mu m = 0.6 \times {10^{ - 6}}m$
Energy that is converted to light$ = 25\% $
The relation between energy and frequency of one photon can be given using Planck’s equation as per which $ = \dfrac{{hc}}{\lambda }$---(i)
Where ‘h’ is Planck’s constant
‘c’ is the speed of light
Substituting the given values in equation (i), we get
$\Rightarrow E = \dfrac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{0.6 \times {{10}^{ - 6}}}}$
$ \Rightarrow E = 33 \times {10^{ - 20}}J$
Energy radiated as yellow light per second is given by
$ \Rightarrow E = 200 \times \dfrac{{25}}{{100}}$
$ \Rightarrow E = 50Watt$
Number of photons emitted per second by yellow light is given by $ = \dfrac{{50}}{{33 \times {{10}^{ - 20}}}}$ $ \approx 1.5 \times {10^{20}}$
The number of photons of yellow light it emits per second is: $1.5 \times {10^{20}}$.
Hence, Option C is the right answer.
Note: It is important to remember that in the phenomenon of photoelectric effect, the particle nature of light is considered. The light is considered as a stream of particles carrying electromagnetic energy. These particles of light are known as photons. These photons have different frequencies and different energies. The condition for photoelectric effect is that the energy of the incident photons should be enough to overcome the attractive forces between electrons on the surface of the metal. This minimum energy acquired by photons is known as threshold energy. If the energy of photons is less than threshold energy, no photoelectric effect will take place.
Complete step by step solution:
Given that $P = 200W$
wavelength$\lambda = 0.6\mu m = 0.6 \times {10^{ - 6}}m$
Energy that is converted to light$ = 25\% $
The relation between energy and frequency of one photon can be given using Planck’s equation as per which $ = \dfrac{{hc}}{\lambda }$---(i)
Where ‘h’ is Planck’s constant
‘c’ is the speed of light
Substituting the given values in equation (i), we get
$\Rightarrow E = \dfrac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{0.6 \times {{10}^{ - 6}}}}$
$ \Rightarrow E = 33 \times {10^{ - 20}}J$
Energy radiated as yellow light per second is given by
$ \Rightarrow E = 200 \times \dfrac{{25}}{{100}}$
$ \Rightarrow E = 50Watt$
Number of photons emitted per second by yellow light is given by $ = \dfrac{{50}}{{33 \times {{10}^{ - 20}}}}$ $ \approx 1.5 \times {10^{20}}$
The number of photons of yellow light it emits per second is: $1.5 \times {10^{20}}$.
Hence, Option C is the right answer.
Note: It is important to remember that in the phenomenon of photoelectric effect, the particle nature of light is considered. The light is considered as a stream of particles carrying electromagnetic energy. These particles of light are known as photons. These photons have different frequencies and different energies. The condition for photoelectric effect is that the energy of the incident photons should be enough to overcome the attractive forces between electrons on the surface of the metal. This minimum energy acquired by photons is known as threshold energy. If the energy of photons is less than threshold energy, no photoelectric effect will take place.
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