(A) According to Wien's displacement law, $\lambda_{max} T = b$, where $b = 2.9 \times 10^{-3} m K$ is Wien's constant.Given $\lambda_{max} = 2900 Å =

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

(A) According to Wien's displacement law, $\lambda_{max} T = b$, where $b = 2.9 \times 10^{-3} m K$ is Wien's constant.Given $\lambda_{max} = 2900 Å = 2900 \times 10^{-10} m = 2.9 \times 10^{-7} m$.Substituting the values, $T = \frac{b}{\lambda_{max}} = \frac{2.9 \times 10^{-3}}{2.9 \times 10^{-7}} = 10^4 K$.The intensity of radiation emitted by a perfect radiator (black body) is given by the Stefan-Boltzmann law: $E = \sigma T^4$.Given $\sigma = 5.67 \times 10^{-8} W m^{-2} K^{-4}$.$E = (5.67 \times 10^{-8}) \times (10^4)^4 = 5.67 \times 10^{-8} \times 10^{16} = 5.67 \times 10^8 W m^{-2}$.Thus, the correct option is $A$.

Class 11 Physics 2 Thermal Properties of Matter: Heat Transfer Radiation by Stefan's Boltzmann Law

Medium

If the radiation emitted by a perfect radiator has maximum intensity at a wavelength of $2900 Å$, the intensity of radiation emitted by it is (Stefan-Boltzmann's constant $= 5.67 \times 10^{-8} W m^{-2} K^{-4}$ and Wien's constant $= 2.9 \times 10^{-3} m K$).

TS EAMCET 2025 All TS EAMCET

A
$5.67 \times 10^8 W m^{-2}$
B
$5.67 W m^{-2}$
C
$5670 W m^{-2}$
D
$2.9 W m^{-2}$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

2 Thermal Properties of Matter: Heat Transfer

Subtopic

Radiation by Stefan's Boltzmann Law

Previous Year

TS EAMCET 2025 Paper All TS EAMCET years →

Three very large plates of same area are kept parallel and close to each other. They are considered as ideal black surfaces and have very high thermal conductivity. First and third plates are maintained at absolute temperatures $2 T$ and $3 T$ respectively. Temperature of the middle plate in steady state is

EasyAP EAMCET 2018

View Solution

Two spheres $P$ and $Q$,of the same color,having radii $8 \; cm$ and $2 \; cm$ are maintained at temperatures $127^{\circ}C$ and $527^{\circ}C$ respectively. The ratio of energy radiated by $P$ and $Q$ is:

Medium

View Solution

$A$ black rectangular surface of area $A$ emits energy $E$ per second at $27^{\circ}C$. If the length and breadth are reduced to half of their initial values and the temperature is raised to $327^{\circ}C$, then the energy emitted per second becomes: (in $E$)

DifficultMHT CET 2026

View Solution

Star $A$ has radius $r$ and surface temperature $T$,while star $B$ has radius $4r$ and surface temperature $T/2$. The ratio of the power radiated by the two stars,$P_A : P_B$,is:

Medium

View Solution

$A$ wire of length $L$ and diameter $d$ is used in a bulb. The temperature of the wire is $T$ and the power radiated by the wire is $P$. Its emissivity is $e$ ($\sigma$ = Stefan's constant). (Assume that the emissivity of the wire material is the same at all wavelengths).

MediumMHT CET 2026

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

If the radiation emitted by a perfect radiator has maximum intensity at a wavelength of $2900 Å$, the intensity of radiation emitted by it is (Stefan-Boltzmann's constant $= 5.67 \times 10^{-8} W m^{-2} K^{-4}$ and Wien's constant $= 2.9 \times 10^{-3} m K$).

Similar Questions

Two spheres $P$ and $Q$,of the same color,having radii $8 \; cm$ and $2 \; cm$ are maintained at temperatures $127^{\circ}C$ and $527^{\circ}C$ respectively. The ratio of energy radiated by $P$ and $Q$ is:

$A$ black rectangular surface of area $A$ emits energy $E$ per second at $27^{\circ}C$. If the length and breadth are reduced to half of their initial values and the temperature is raised to $327^{\circ}C$, then the energy emitted per second becomes: (in $E$)

Star $A$ has radius $r$ and surface temperature $T$,while star $B$ has radius $4r$ and surface temperature $T/2$. The ratio of the power radiated by the two stars,$P_A : P_B$,is:

$A$ wire of length $L$ and diameter $d$ is used in a bulb. The temperature of the wire is $T$ and the power radiated by the wire is $P$. Its emissivity is $e$ ($\sigma$ = Stefan's constant). (Assume that the emissivity of the wire material is the same at all wavelengths).

Vedclass Test Series

Exam Paper Generator

Online Exam Module