If the Kelvin temperature of a body is doubled, the wavelength of the
most intense light is...
a. 1/16
b. half
c. 1/2
d. 2 times
e. 16 times
...the wavelength originally of the most intense light

Hello!!

We can solve this using the Wien's law.

Wien's law in physics states that there is an inverse relationship
between the wavelength of the peak of the emission of a body (or
sometimes Blackbody Radiation for arcane physical reasons) and its
temperature.

ëmax = (0.002898 m) / T 
where T is the temperature of the blackbody in kelvin (K) and ëmax is
the peak wavelength.
Basically, the hotter an object is, the shorter the wavelength at
which it will emit radiation.


Then if you double the Kelvin temperature, then the new wavelenght of
the most intense light is is half of the original one.

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Clarification of Answer by livioflores-ga on 06 Dec 2002 22:49 PST

ëmax is equal to Wavelenght max.

For reference see:
"Wien's law - From Wikipedia, the free encyclopedia"
http://www.wikipedia.org/wiki/Wien's_law


"Wien's Law"
http://www.astro.umd.edu/education/astro/sprop/wien.html


"Wien's Law"
http://astrosun.tn.cornell.edu/courses/astro201/wiens_law.htm


Search Keywords :
Thermal Radiation
Wien's law

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If you need some clarification, please post a request for it.

Thank you

livioflores-ga

---

Clarification of Answer by livioflores-ga on 06 Dec 2002 23:55 PST

ëmax is equal to Wavelenght max (Wmax);
Then Wmax = (0.002898 m)/ T

Initial Wmax = (0.002898 m)/ Ti
Final Wmax = (0.002898 m)/ 2Ti = 1/2 x (0.002898 m)/ Ti = 1/2 x Iniyial Wmax

May be this clarify.

livioflores-ga