
The light from the sun is found to have a maximum intensity near the wavelength of $470\;{\text{nm}}$. Assuming the surface of the sun as a black body, the temperature of the sun is …………………..
[ Wien’s constant $b = 2.898 \times {10^{ - 3}}\;{\text{mK}}$]
A. $5800\;{\text{K}}$
B. $6050\;{\text{K}}$
C. $6166\;{\text{K}}$
D. $6500\;{\text{K}}$
Answer
232.8k+ views
Hint The above problem is based on Wien's displacement law. This law gives the relationship between the maximum wavelengths of the light that can be emitted from the source. The product of the wavelength of the light and temperature of the light is constant and this constant is known as the Wien’s constant.
Complete step by step answer
Given: The wavelength of the light for maximum intensity is ${\lambda _{\max }} = 470\;{\text{nm}} = 470\;{\text{nm}} \times \dfrac{{{{10}^{ - 9}}\;{\text{m}}}}{{1\;{\text{nm}}}} = 4.70 \times {10^{ - 7}}\;{\text{m}}$.
The Wien’s constant is $b = 2.898 \times {10^{ - 3}}\;{\text{mK}}$.
Apply the Wien’s displacement law to calculate the formula for temperature of the sun.
${\lambda _{\max }}T = b$
Substitute $2.898 \times {10^{ - 3}}\;{\text{mK}}$ for b and $4.70 \times {10^{ - 7}}\;{\text{m}}$ for ${\lambda _{\max }}$ in the above expression to find the temperature of the sun.
$\left( {4.70 \times {{10}^{ - 7}}\;{\text{m}}} \right)T = 2.898 \times {10^{ - 3}}\;{\text{mK}}$
$T = 6165.95\;{\text{K}}$
$T \approx 6166\;{\text{K}}$
Thus, the temperature of the sun is $6166\;{\text{K}}$ and the option (C) is the correct answer.
Additional information The thermal radiation emitted by a body becomes visible at high temperatures. The maximum temperature of the radiation becomes observable for shorter wavelengths. The temperature of the body is inversely proportional to the wavelength of the emitted thermal radiation. The area under the curve between the wavelength and temperature of the body gives the total emissive power of the body.
Note Conversion of temperature in Kelvin from degree Celsius and unit of wavelength in meters is necessary before applying the Wien’s displacement law. The body emits the higher thermal radiation for a smaller wavelength of the light.
Complete step by step answer
Given: The wavelength of the light for maximum intensity is ${\lambda _{\max }} = 470\;{\text{nm}} = 470\;{\text{nm}} \times \dfrac{{{{10}^{ - 9}}\;{\text{m}}}}{{1\;{\text{nm}}}} = 4.70 \times {10^{ - 7}}\;{\text{m}}$.
The Wien’s constant is $b = 2.898 \times {10^{ - 3}}\;{\text{mK}}$.
Apply the Wien’s displacement law to calculate the formula for temperature of the sun.
${\lambda _{\max }}T = b$
Substitute $2.898 \times {10^{ - 3}}\;{\text{mK}}$ for b and $4.70 \times {10^{ - 7}}\;{\text{m}}$ for ${\lambda _{\max }}$ in the above expression to find the temperature of the sun.
$\left( {4.70 \times {{10}^{ - 7}}\;{\text{m}}} \right)T = 2.898 \times {10^{ - 3}}\;{\text{mK}}$
$T = 6165.95\;{\text{K}}$
$T \approx 6166\;{\text{K}}$
Thus, the temperature of the sun is $6166\;{\text{K}}$ and the option (C) is the correct answer.
Additional information The thermal radiation emitted by a body becomes visible at high temperatures. The maximum temperature of the radiation becomes observable for shorter wavelengths. The temperature of the body is inversely proportional to the wavelength of the emitted thermal radiation. The area under the curve between the wavelength and temperature of the body gives the total emissive power of the body.
Note Conversion of temperature in Kelvin from degree Celsius and unit of wavelength in meters is necessary before applying the Wien’s displacement law. The body emits the higher thermal radiation for a smaller wavelength of the light.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, 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

Understanding Uniform Acceleration in Physics

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

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

Waves Class 11 Physics Chapter 14 CBSE Notes - 2025-26

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

