
If mean wavelength of light radiated by 100 W lamp is \[5000\mathop A\limits^0 \], then number of photons radiated per second are
A. \[3 \times {10^{23}}\]
B. \[2.5 \times {10^{22}}\]
C. \[2.5 \times {10^{20}}\]
D. \[5 \times {10^{17}}\]
Answer
219.6k+ views
Hint: In this question we need to calculate the number of photons radiated per second. In order to find it, we first need to calculate the energy emitted by the photons that is converted into light by using the formula. Then, we can calculate the energy of n photons. Based on the relation between energy, power and time we can find the result.
Formula usedFormula for the power is given as,
\[P = \dfrac{\text{Work(W)}}{\text{Time(t)}} = \dfrac{\text{Energy(E)}}{\text{Time(t)}}\]
The energy emitted by the photons is given as,
\[E = \dfrac{{hc}}{\lambda }\]
Where h is Planck constant, c is the speed of light and \[\lambda \] is the wavelength.
Complete step by step solution:
Mean wavelength of light, \[\lambda = 5000\mathop A\limits^0 \]
Power, \[P = 100W\]
As we know,
\[\text{Power, P} = \dfrac{\text{Work(W)}}{\text{Time(t)}} = \dfrac{\text{Energy(E)}}{\text{Time(t)}}\]
If the energy of 1 photon is E.
Then the energy of $n$ photon is \[nE\].
\[P = \dfrac{{nE}}{t}\]
Using the formula for E,
\[P = \left( {\dfrac{n}{t}} \right)\dfrac{{hc}}{\lambda }\]
\[\dfrac{n}{t} = \dfrac{{6.62 \times {{10}^{34}}}}{{5000 \times {{10}^{ - 10}}}}\]
By solving this, we get
\[\dfrac{n}{t} = 2.5 \times {10^{20}}\]
Therefore, the number of photons radiated per second is \[2.5 \times {10^{20}}\].
Hence option C is the correct answer.
Note: The energy of a single photon emitted by the lamp can be determined by the energy formula. We are able to calculate the number of photons by simply dividing the total energy by the energy of a single photon due to the fact that electromagnetic radiation is always quantized in nature. This implies that the lamp emits light in packets of fixed energy known as photons. In physics the power is defined as the amount of energy transferred or converted from one form to the another in per unit time.
Formula usedFormula for the power is given as,
\[P = \dfrac{\text{Work(W)}}{\text{Time(t)}} = \dfrac{\text{Energy(E)}}{\text{Time(t)}}\]
The energy emitted by the photons is given as,
\[E = \dfrac{{hc}}{\lambda }\]
Where h is Planck constant, c is the speed of light and \[\lambda \] is the wavelength.
Complete step by step solution:
Mean wavelength of light, \[\lambda = 5000\mathop A\limits^0 \]
Power, \[P = 100W\]
As we know,
\[\text{Power, P} = \dfrac{\text{Work(W)}}{\text{Time(t)}} = \dfrac{\text{Energy(E)}}{\text{Time(t)}}\]
If the energy of 1 photon is E.
Then the energy of $n$ photon is \[nE\].
\[P = \dfrac{{nE}}{t}\]
Using the formula for E,
\[P = \left( {\dfrac{n}{t}} \right)\dfrac{{hc}}{\lambda }\]
\[\dfrac{n}{t} = \dfrac{{6.62 \times {{10}^{34}}}}{{5000 \times {{10}^{ - 10}}}}\]
By solving this, we get
\[\dfrac{n}{t} = 2.5 \times {10^{20}}\]
Therefore, the number of photons radiated per second is \[2.5 \times {10^{20}}\].
Hence option C is the correct answer.
Note: The energy of a single photon emitted by the lamp can be determined by the energy formula. We are able to calculate the number of photons by simply dividing the total energy by the energy of a single photon due to the fact that electromagnetic radiation is always quantized in nature. This implies that the lamp emits light in packets of fixed energy known as photons. In physics the power is defined as the amount of energy transferred or converted from one form to the another in per unit time.
Recently Updated Pages
Electricity and Magnetism Explained: Key Concepts & Applications

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

States of Matter Chapter For JEE Main Chemistry

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Understanding Uniform Acceleration in Physics

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Atomic Structure for Beginners

Other Pages
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

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Centrifugal Force in Physics

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Degree of Dissociation: Meaning, Formula, Calculation & Uses

