Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 59

If a blackbody is radiating at \(T=1650 \mathrm{K},\) at what wavelength is the maximum intensity?

Short Answer

Expert verified
Answer: The maximum intensity for a blackbody at a temperature of 1650 K is radiated at a wavelength of approximately 1.756 x 10^-6 m or 1756 nm.

Step by step solution

01

Understand Wien's Displacement Law

Wien's Displacement Law formula states that the product of the maximum wavelength (λ_max) and the temperature (T) of a blackbody is a constant, represented by the symbol b: λ_max * T = b Here, b is Wien's displacement constant, which is equal to 2.898 x 10^-3 m*K.
02

Find the maximum wavelength (λ_max)

Now that we have the temperature (T) and Wien's constant (b), we can find the maximum wavelength (λ_max) as follows: λ_max = b / T Now, plug in the values for b (2.898 x 10^-3 m*K) and T (1650 K): λ_max = (2.898 x 10^-3 m*K) / (1650 K)
03

Calculate the result

After dividing the values, we get: λ_max = 1.756 x 10^-6 m This value is the maximum wavelength at which the maximum intensity is radiated by the blackbody. So, the maximum intensity for a blackbody at a temperature of 1650 K is radiated at a wavelength of approximately 1.756 x 10^-6 m or 1756 nm.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The inner vessel of a calorimeter contains \(2.50 \times 10^{2} \mathrm{g}\) of tetrachloromethane, \(\mathrm{CCl}_{4},\) at \(40.00^{\circ} \mathrm{C} .\) The vessel is surrounded by \(2.00 \mathrm{kg}\) of water at $18.00^{\circ} \mathrm{C} .\( After a time, the \)\mathrm{CCl}_{4}$ and the water reach the equilibrium temperature of \(18.54^{\circ} \mathrm{C} .\) What is the specific heat of \(\mathrm{CCl}_{4} ?\)
A mass of \(1.4 \mathrm{kg}\) of water at \(22^{\circ} \mathrm{C}\) is poured from a height of \(2.5 \mathrm{m}\) into a vessel containing \(5.0 \mathrm{kg}\) of water at \(22^{\circ} \mathrm{C} .\) (a) How much does the internal energy of the \(6.4 \mathrm{kg}\) of water increase? (b) Is it likely that the water temperature increases? Explain.
A birch tree loses \(618 \mathrm{mg}\) of water per minute through transpiration (evaporation of water through stomatal pores). What is the rate of heat lost through transpiration?
A metal rod with a diameter of \(2.30 \mathrm{cm}\) and length of $1.10 \mathrm{m}\( has one end immersed in ice at \)32.0^{\circ} \mathrm{F}$ and the other end in boiling water at \(212^{\circ} \mathrm{F}\). If the ice melts at a rate of 1.32 g every 175 s, what is the thermal conductivity of this metal? Identify the metal. Assume there is no heat lost to the surrounding air.
Bare, dark-colored basalt has a thermal conductivity of 3.1 $\mathrm{W} /(\mathrm{m} \cdot \mathrm{K}),$ whereas light-colored sandstone's thermal conductivity is only \(2.4 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K}) .\) Even though the same amount of radiation is incident on both and their surface temperatures are the same, the temperature gradient within the two materials will differ. For the same patch of area, what is the ratio of the depth in basalt as compared with the depth in sandstone that gives the same temperature difference?
See all solutions

Recommended explanations on Physics Textbooks

Scientific Method Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Engineering Physics

Read Explanation

Measurements

Read Explanation

Physical Quantities And Units

Read Explanation

Solid State Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free