Question

what is khirchoof law​

Answer 1

$\mathcal{KIRCHHOFF'S\:\:LAW}$

$\implies$ The ratio of emissive power to absorptive power is same for all surfaces at the same temperature and is equal to the emissive power of a perfectly black body at that temperature.

Experimentally a very good approximation to a black body is provided by a cavity enclosed by a high temperature opaque walls.

Thus, if $\sf{a_{practical}}$ and $\sf{E_{practical}}$ represent the absoptive and emissive power of a given surface, while $\sf{a_{black}}$ and $\sf{E_{black}}$ for a perfectly black body, then according to law

$\sf{\dfrac{E_{practical}}{a_{practical}}=\dfrac{E_{black}}{a_{black}}}$

But for a perfectly black body, $\sf{a_{black}=1}$

So $\sf{\dfrac{E_{practical}}{a_{practical}}=E_{black}}$

If emissive and absorptive powers are considered for a particular wavelength $\lambda$,

$\sf{\left(\dfrac{E_{\lambda}}{a_{\lambda}}\right)_{practical}=\left(E_{\lambda}\right)_{black}}$

Now since $\sf{\left(E_{\lambda}\right)_{black}}$ is constant at a given temperature, according to this law if a surface is a good absorber of a particular wavelength it is also a good emitter of that wavelength.

This in turn implies that a good absorber is a good emitter.

Answer 2

IRCHHOFF

′

SLAW

\implies⟹ The ratio of emissive power to absorptive power is same for all surfaces at the same temperature and is equal to the emissive power of a perfectly black body at that temperature.

Experimentally a very good approximation to a black body is provided by a cavity enclosed by a high temperature opaque walls.

Thus, if \sf{a_{practical}}a

practical

and \sf{E_{practical}}E

practical

represent the absoptive and emissive power of a given surface, while \sf{a_{black}}a

black

and \sf{E_{black}}E

black

for a perfectly black body, then according to law

\sf{\dfrac{E_{practical}}{a_{practical}}=\dfrac{E_{black}}{a_{black}}}

a

practical

E

practical

=

a

black

E

black

But for a perfectly black body, \sf{a_{black}=1}a

black

=1

So \sf{\dfrac{E_{practical}}{a_{practical}}=E_{black}}

a

practical

E

practical

=E

black

If emissive and absorptive powers are considered for a particular wavelength \lambdaλ ,

\sf{\left(\dfrac{E_{\lambda}}{a_{\lambda}}\right)_{practical}=\left(E_{\lambda}\right)_{black}}(

a

λ

E

λ

)

practical

=(E

λ

)

black

Now since \sf{\left(E_{\lambda}\right)_{black}}(E

λ

)

black

is constant at a given temperature, according to this law if a surface is a good absorber of a particular wavelength it is also a good emitter of that wavelength.

This in turn implies that a good absorber is a good emitter.