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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Thermodynamics

Light from a source has its maximum intensity near the wavelength 500 nm . Assuming the body to be a black body calculate the temperature at its surface ?

$(a)\;5760\;K\qquad(b)\;2880\;K\qquad(c)\;1440\;K\qquad(d)\;cant\;be\;determined$

1 Answer

Answer : $5760 K$
Explanation :
For a blackbody , $\;\lambda T=constant = b \;$ (wien's displacement law)
Since $\;b=2.88 \times 10^{-3} mK$
$T=\large\frac{2.88 \times 10^{-3}}{500 \times 10^{-9}}=5760 K$
answered Mar 13, 2014 by yamini.v
 

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