
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
225.6k+ 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.
Recently Updated Pages
JEE Main 2025-26 Experimental Skills Mock Test – Free Practice

JEE Main 2025-26: Magnetic Effects of Current & Magnetism Mock Test

JEE Main 2025-26 Atoms and Nuclei Mock Test – Free Practice Online

JEE Main Mock Test 2025-26: Optics Chapter Practice Online

The work done in slowly moving an electron of charge class 12 physics JEE_Main

The value of the resistor RS needed in the DC voltage class 12 physics JEE_Main

Trending doubts
JEE Main 2026: City Intimation Slip and Exam Dates Released, Application Form Closed, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

Dual Nature of Radiation and Matter Class 12 Physics Chapter 11 CBSE Notes - 2025-26

Understanding Atomic Structure for Beginners

Understanding Electromagnetic Waves and Their Importance

