Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

At what wavelength does the human body emit the maximum electromagnetic radiation? Use Wien's law from Exercise 14 and assume a skin temperature of \(70^{\circ} \mathrm{F}\).

Short Answer

Expert verified
The wavelength at which the human body (with a skin temperature of \(70^{\circ} \mathrm{F}\)) emits the maximum electromagnetic radiation can be found using Wien's Law after converting the temperature from Fahrenheit to Kelvin. The computation involves substituting the known values into the formula.

Step by step solution

01

Convert Fahrenheit to Kelvin

To change temperature from Fahrenheit to Kelvin, the formula to be used is \[T(K) = \frac{5}{9}(T(F) - 32) + 273.15\] So the temperature in Kelvin is: \[T(K) = \frac{5}{9}(70 - 32) + 273.15\]
02

Apply Wien's Law

Once the temperature in Kelvin is obtained, Wien's Law can be applied. The formula is \[λ_{max} = \frac{b}{T}\], where b is the Wien's constant equal to \(2.898 \times 10^{-3} m.K\). Substitute the values for T(K) obtained in step 1 and b into the equation to find \(λ_{max}\).
03

Compute \(λ_{max}\)

Finally, after substituting the values into the Wien's Law formula, compute the result which will give the maximum wavelength \(λ_{max}\) at which the human body emits the maximum radiation.

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.

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

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electromagnetic Radiation
Electromagnetic radiation is a form of energy that spreads out as it travels through space. It can be visualized as waves composed of both electric and magnetic fields oscillating perpendicular to each other, moving at the speed of light. This radiation includes a vast spectrum of wavelengths and frequencies, from very long radio waves to very short gamma rays. The human body itself emits infrared radiation because of the heat it produces. Understanding electromagnetic radiation is crucial in many fields, including astronomy, telecommunications, and medicine.

For instance, when discussing the heat from a human body or the light emitted by a star, we are referring to types of electromagnetic radiation. This also relates directly to Wien’s Law, which connects the temperature of an object to the peak wavelength of the radiation it emits.
Blackbody Radiation
Blackbody radiation refers to a hypothetical object that is a perfect emitter and absorber of all forms of electromagnetic radiation. In reality, no perfect blackbodies exist, but many objects act approximately like blackbodies. For example, stars, including our sun, can be treated as blackbodies to simplify the study of the radiation they emit.

Blackbody radiation is characterized by a spectrum that depends only on the object's temperature. An important aspect of this kind of radiation is its predictability, as described by Wien's Law. This allows us to estimate the temperature of an object based on the radiation it emits, which is invaluable in understanding the physical properties of stars and other astronomical objects.
Temperature Conversion
Temperature conversion is crucial when working with different temperature scales. The most commonly used scales are Celsius, Fahrenheit, and Kelvin. Each scale has its own applications; for example, Celsius is widely used in most of the world for daily weather reports, while Fahrenheit is primarily used in the United States. The Kelvin scale, however, is the standard unit of measure for temperature in the sciences.

To convert Fahrenheit to Kelvin, which is necessary for applying Wien’s Law, you can use the conversion formula \(T(K) = \frac{5}{9}(T(F) - 32) + 273.15\). It is important to make this conversion because Wien's Law operates with temperatures measured in Kelvin only. Failing to correctly convert temperatures can lead to inaccurate calculations of emitted radiation wavelengths.
Kelvin Scale
The Kelvin scale is an absolute temperature scale that measures temperature from absolute zero, theoretically the coldest possible temperature, where molecular movement stops. One Kelvin represents the same temperature change as one degree Celsius, but they have different starting points: 0K is -273.15°C. Unlike Celsius and Fahrenheit, the Kelvin scale does not utilize the term 'degree' and is written simply as 'Kelvin' (K).

In scientific contexts, especially in physics and astronomy, the Kelvin scale is used because many formulas, including Wien's Law, are based on absolute temperature measurements. This scale eliminates the negative temperature values that can complicate calculations, making it a standardized unit for scientific research.

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

A photon and an object of mass \(m\) have the same momentum \(p\). (a) Assuming that the massive object is moving slowly, so that nonrelativistic fonnulas are valid. find in terms of \(m, p\), and \(c\) the ratio of the massive ob)ect's kinetic energy to the photon's kinetic ener \(g y\), and argue that it is small. (b) Find the same ratio found in part (a), but using relativistically correct formulas for the massive object. (Nore: \(E^{2}=p^{2} c^{2}+m^{2} c^{4}\) may be helpful.) (c) Show that the low-speed limit of the ratio of part (b) agrees with part (a) and that the high-speed limit is \(1 .\) (d) Show that at very high speod, the kinetic energy of a massive object approaches \(\rho \mathrm{c}\),

Suppose we produce X-rays not by smashing elections into targets but by smashing protons, which are far more massive. If the same accelerating potential difference were used for both, how would the cutoff wavelengths of the two X-ray spectra compare? Explain.

A low-intensity beam of light is sent toward a narrow single slit. On the far side, individual flashes are seen sporadically at detectors over a broad area that is orders of magnitude wider than the slit width. What aspects of the experiment suggest a wave nature for light, and what aspects suggest a particle nature?

We analyze the photoelectric effect using photon energy alone. Why isn't the photon momentum a consideration? (It may help to reread the discussion of momentum and energy in connection with pair production.)

A function \(f(\alpha)\) is nonzero only in the region of width \(2 \delta\) centered at \(\alpha=0\) $$ f(\alpha)=\left\\{\begin{array}{ll} C & |\alpha| \leq \delta \\ 0 & |\alpha|>\delta \end{array}\right. $$ where \(C\) is a constant. (a) Find and plot versus \(\beta\) the Fourier transform \(A(\beta)\) of this function. (b) The function \(f \alpha\) ) might represent a pulse occupying either finite distance \((\alpha=\) position) or finite time \((\alpha\) = time). Comment on the wave number spectrum if \(\alpha\) is position and on the frequency spectrum if \(\alpha\) is time. Specifically address the dependence of the width of the spectrum on \(\delta\).

See all solutions

Recommended explanations on Physics Textbooks

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