
A star which appears blue will be
A) much hotter than the Sun
B) colder than the Sun
C) as hot as the Sun
D) at $ - {273^ \circ }C$
Answer
139.8k+ views
Hint: Energy is inversely proportional to the wavelength and is directly proportional to the temperature of the body. So just check the relation and compare temperature. The light colour with less wavelength will have more temperature.
Complete step by step answer:
Explanation of the symbols used:
$h$ = Planck’s constant
$c$ = speed of light
${E_{star}}$ = energy per unit area (flux) radiated by the blue star
${E_{sun}}$ = energy per unit area (flux) radiated by the Sun
$\sigma $ = Stefan-Boltzmann constant
$\varepsilon $ = emissivity of the body
${T_{star}}$ = temperature of the blue star
${T_{sun}}$ = temperature of the Sun
We know that energy $E$ is inversely proportional to the wavelength $\lambda $ according to the relation
$E = \dfrac{{hc}}{\lambda }$.
Hence as we know the wavelength of blue light is less than the wavelength of yellow or red light (colour of the Sun). The energy of the blue star is therefore more than the energy of the Sun (since it is inversely proportional).
Therefore,
${E_{star}} > {E_{sun}}$
Hence according to Stefan-Boltzmann Law we know that $E = \sigma \varepsilon {T^4}$
$\sigma \varepsilon T_{star}^4 > \sigma \varepsilon T_{sun}^4$
${T_{star}} > T_{sun}^{}$
Therefore, the temperature of the star will be greater than the temperature of the Sun, i.e., the star which appears blue will be much hotter than the Sun.
Hence the correct answer is option A.
Note: The comparison can be made by Wien’s displacement law as well which directly relates the wavelength of emitted radiation and temperature of a body as
$\lambda T$ = constant
Which means since blue light has less wavelength as compared to Yellow or Orange light, it will have more temperature, i.e., it will be hotter. This method can be used since it is a shorter method but our solution that we have discussed is more fundamental in nature and hence generally recommended.
Complete step by step answer:
Explanation of the symbols used:
$h$ = Planck’s constant
$c$ = speed of light
${E_{star}}$ = energy per unit area (flux) radiated by the blue star
${E_{sun}}$ = energy per unit area (flux) radiated by the Sun
$\sigma $ = Stefan-Boltzmann constant
$\varepsilon $ = emissivity of the body
${T_{star}}$ = temperature of the blue star
${T_{sun}}$ = temperature of the Sun
We know that energy $E$ is inversely proportional to the wavelength $\lambda $ according to the relation
$E = \dfrac{{hc}}{\lambda }$.
Hence as we know the wavelength of blue light is less than the wavelength of yellow or red light (colour of the Sun). The energy of the blue star is therefore more than the energy of the Sun (since it is inversely proportional).
Therefore,
${E_{star}} > {E_{sun}}$
Hence according to Stefan-Boltzmann Law we know that $E = \sigma \varepsilon {T^4}$
$\sigma \varepsilon T_{star}^4 > \sigma \varepsilon T_{sun}^4$
${T_{star}} > T_{sun}^{}$
Therefore, the temperature of the star will be greater than the temperature of the Sun, i.e., the star which appears blue will be much hotter than the Sun.
Hence the correct answer is option A.
Note: The comparison can be made by Wien’s displacement law as well which directly relates the wavelength of emitted radiation and temperature of a body as
$\lambda T$ = constant
Which means since blue light has less wavelength as compared to Yellow or Orange light, it will have more temperature, i.e., it will be hotter. This method can be used since it is a shorter method but our solution that we have discussed is more fundamental in nature and hence generally recommended.
Recently Updated Pages
Difference Between Circuit Switching and Packet Switching

Difference Between Mass and Weight

JEE Main Participating Colleges 2024 - A Complete List of Top Colleges

JEE Main Maths Paper Pattern 2025 – Marking, Sections & Tips

Sign up for JEE Main 2025 Live Classes - Vedantu

JEE Main 2025 Helpline Numbers - Center Contact, Phone Number, Address

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Other Pages
Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

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

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
