Ratio of energy densities of black body radiation

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

The energy density is given as $u = \frac { 8 \pi {\nu }^2}{c^3}~ \frac { h \nu} {e^{ \frac { h\nu}{k_B T}} – 1}.$

EDIT : I put the constant C.
$\frac { u( 2 \nu) } {u(\nu)} = C \frac { {e^{ \frac { h\nu}{k_B T}} – 1} }{ {e^{ \frac { 2h\nu}{k_B T}} – 1} }$
Where C is the appropriate constant.

Is this correct?

 

[ehild]

Homework Statement

View attachment 217649

Homework Equations

The Attempt at a Solution

The energy density is given as $u = \frac { 8 \pi {\nu }^2}{c^3}~ \frac { h \nu} {e^{ \frac { h\nu}{k_B T}} – 1}$.

$\frac { u( 2 \nu) } {u(\nu)} = \frac { {e^{ \frac { h \nu}{k_B T}} – 1} }{ {e^{ \frac { 2h \nu}{k_B T}} – 1} }$

Is this correct?

The formula is correct (apart from a constant factor), but you have to simplify it and figure out which option it corresponds to. Note that $e^{ \frac { 2h \nu}{k_B T}}=\left(e^{ \frac { h \nu}{k_B T}}\right)^2$

 

[Pushoam]

Note that $e^{ \frac { 2h \nu}{k_B T}}=\left(e^{ \frac { h \nu}{k_B T}}\right)^2$

Thanks for this insight.
$\frac { u( 2 \nu) } {u(\nu)} = \frac { {e^{ \frac { h\nu}{k_B T} } – 1} } { \left ({e^{ \frac { h\nu}{k_B T}} – 1}\right ) \left( {e^{ \frac { h\nu}{k_B T}} +1} \right) }$ $= \frac1 {e^{ \frac { h\nu}{k_B T}} +1}$

So, the answer is option (B).

 

[ehild]

Thanks for this insight.
$\frac { u( 2 \nu) } {u(\nu)} = \frac { {e^{ \frac { h\nu}{k_B T} } – 1} } { \left ({e^{ \frac { h\nu}{k_B T}} – 1}\right ) \left( {e^{ \frac { h\nu}{k_B T}} +1} \right) }$ $= \frac1 {e^{ \frac { h\nu}{k_B T}} +1}$

So, the answer is option (B).

Yes :)