 # Recall that light is EM radiation and is therefore characterized by its wavelength. Recall that light is EM radiation and is therefore characterized by.

## Presentation on theme: "Recall that light is EM radiation and is therefore characterized by its wavelength. Recall that light is EM radiation and is therefore characterized by."— Presentation transcript:

Recall that light is EM radiation and is therefore characterized by its wavelength. Recall that light is EM radiation and is therefore characterized by its wavelength.

Imagine we had the “perfect” emitter of EM radiation. We would call that a “Blackbody”. The Sun (and all stars) give off nearly every wavelength of EM radiation. But what wavelength dominates????? During this Solar Eclipse (1991) you can see the Corona of the Sun. It is extends hundreds of thousands of Kilometers into space

The simplest and most common way to produce EM radiation is to heat up an object. The simplest and most common way to produce EM radiation is to heat up an object.

Lets consider the welder here. Lets consider the welder here. As the temperature increases what can we say about... As the temperature increases what can we say about... The color? The amount of Heat?

BLACKBODY CURVES

The Blackbody curve for the Sun The Blackbody curve for the Sun When we compare the curve for a star to the curve of a blackbody at a given temp, we can in effect... When we compare the curve for a star to the curve of a blackbody at a given temp, we can in effect... Take the temperature of a star from millions/billions of miles away!!!

The equations that describe these curves are rather complicated. The equations that describe these curves are rather complicated. However, we can deduce 2 useful expressions. The first one relates temperature to wavelength of maximum emission. However, we can deduce 2 useful expressions. The first one relates temperature to wavelength of maximum emission. aka Wien’s Displacement Law (PSRT)

Example: Determine the Surface Temperature of the Sun. Example: Determine the Surface Temperature of the Sun. Assume λ max is 500 nm Assume λ max is 500 nm (PSRT)

Thought Experiment. Consider a hot metal rod and a burning match at the same temp. Which gives off more energy??? We must consider the amount of energy emitted per surface area. ENERGY FLUX

The energy flux is just a fancy way of saying The energy flux is just a fancy way of saying “The amount of energy emitted from 1m 2 of an object’s surface per second.” [J/m 2 /sec] or [W/m 2 ] So for the match and metal bar, which one has a higher energy flux? Neither. But clearly the match will not do as much damage to your skin. So for the match and metal bar, which one has a higher energy flux? Neither. But clearly the match will not do as much damage to your skin. So what’s the difference? So what’s the difference?

In 1879, Josef Stefan showed that Five years later Ludwig Boltzmann derived the coefficient that related the 2 quantities. Recall Luminosity is the power output of a star. Therefore....

Sample problem: F1 (c) M02 exam Antares A is part of a binary star system. The companion star Antares B has a surface temperature of 15 000 K and a luminosity that is 1/40 of that of Antares A. Calculate the ratio of the radius of Antares A to Antares B. Antares A has a surface temperature of 3000 K. This info was provided in an earlier part of the problem (not stated). In addition, you must know the formula for the surface area of a sphere to solve these types of problems.

Worksheet on Worksheet on Apparent Brightness Apparent Brightness Wien’s Law Wien’s Law SB Law SB Law

Download ppt "Recall that light is EM radiation and is therefore characterized by its wavelength. Recall that light is EM radiation and is therefore characterized by."

Similar presentations