
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
242.7k+ 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
WBJEE 2026 Registration Started: Important Dates Eligibility Syllabus Exam Pattern

JEE Main 2025-26 Mock Tests: Free Practice Papers & Solutions

JEE Main 2025-26 Experimental Skills Mock Test – Free Practice

JEE Main 2025-26 Electronic Devices Mock Test: Free Practice Online

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

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

Trending doubts
JEE Main 2026: Session 1 Results Out and Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Electric Field of a Uniformly Charged Ring

Other Pages
CBSE Class 12 Physics Question Paper 2026: Download SET-wise PDF with Answer Key & Analysis

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

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

CBSE Class 10 Sanskrit Set 4 52 Question Paper 2025 – PDF, Solutions & Analysis

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

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

