(A) Intensity of radiation is defined as the energy incident per unit area per unit time: $I = \frac{E}{A \cdot t}$.For a photon of frequency $f$, the

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(A) Intensity of radiation is defined as the energy incident per unit area per unit time: $I = \frac{E}{A \cdot t}$.For a photon of frequency $f$, the energy is $E = hf$. Thus, the total energy incident is $E_{total} = N \cdot hf$.For surface $A$: $I_A = \frac{n \cdot hf}{A \cdot t}$.For surface $B$: $I_B = \frac{(2n) \cdot h(3f)}{A \cdot (4t)} = \frac{6nhf}{4At} = \frac{3nhf}{2At}$.Taking the ratio: $\frac{I_A}{I_B} = \frac{nhf}{At} \cdot \frac{2At}{3nhf} = \frac{2}{3}$.Therefore, the ratio is $2: 3$.

Class 12 Physics Dual Nature of Radiation and Matter Particle Nature of Light : Photon

Medium

Consider two black bodies $A$ and $B$ having equal surface areas. On the surface of $A$, $n$ photons of frequency $f$ are incident perpendicularly in a time $t$. On the surface of $B$, $2n$ photons of frequency $3f$ are incident perpendicularly in a time $4t$. The ratio of average intensity of radiation on surface $A$ to that on surface $B$ is

TS EAMCET 2023 All TS EAMCET

A
$2: 3$
B
$3: 2$
C
$1: 12$
D
$1: 24$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Physics Questions

Chapter

Dual Nature of Radiation and Matter

Subtopic

Particle Nature of Light : Photon

Previous Year

TS EAMCET 2023 Paper All TS EAMCET years →

If we express the energy of a photon in $KeV$ and the wavelength in $\mathring{A}$,then the energy of a photon can be calculated from the relation:

Medium

View Solution

The frequency of a photon with energy $100 \ eV$ is ........... $Hz$. (Given: $h = 6.62 \times 10^{-34} \ J \cdot s$,$1 \ eV = 1.6 \times 10^{-19} \ J$)

Easy

View Solution

$A$ source $S_1$ is producing $10^{15}$ photons per second of wavelength $5000 \; \mathring{A}$. Another source $S_2$ is producing $1.02 \times 10^{15}$ photons per second of wavelength $5100 \; \mathring{A}$. Then,$(\text{power of } S_2) / (\text{power of } S_1)$ is equal to

MediumAIPMT 2010

View Solution

Light of wavelength $5000 \, \mathring{A}$ falls on a sensitive surface. If the surface has received $10^{-7} \, \text{J}$ of energy,then the number of photons falling on the surface will be:

Medium

View Solution

$A$ radio transmitter operates at a frequency of $880 \ kHz$ and a power of $10 \ kW$. The number of photons emitted per second is .......

Difficult

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Similar Questions

If we express the energy of a photon in $KeV$ and the wavelength in $\mathring{A}$,then the energy of a photon can be calculated from the relation:

The frequency of a photon with energy $100 \ eV$ is ........... $Hz$. (Given: $h = 6.62 \times 10^{-34} \ J \cdot s$,$1 \ eV = 1.6 \times 10^{-19} \ J$)

$A$ source $S_1$ is producing $10^{15}$ photons per second of wavelength $5000 \; \mathring{A}$. Another source $S_2$ is producing $1.02 \times 10^{15}$ photons per second of wavelength $5100 \; \mathring{A}$. Then,$(\text{power of } S_2) / (\text{power of } S_1)$ is equal to

Light of wavelength $5000 \, \mathring{A}$ falls on a sensitive surface. If the surface has received $10^{-7} \, \text{J}$ of energy,then the number of photons falling on the surface will be:

$A$ radio transmitter operates at a frequency of $880 \ kHz$ and a power of $10 \ kW$. The number of photons emitted per second is .......

Vedclass Test Series

Exam Paper Generator

Online Exam Module