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Untuk panjang gelombang yang besar berlaku pendekatan<br />

exp(hυ/k B T)=exp[hc/(λ k B T)] ≈1+ hυ /k B T<br />

2<br />

8πν<br />

hν<br />

E( ν ) =<br />

3 υ / k T<br />

c<br />

e h B<br />

−1<br />

=<br />

8πν<br />

3<br />

c<br />

2<br />

k<br />

B<br />

T<br />

persamaan dari Raleigh-Jeans.<br />

Persamaan dapat diungkapkan dalam λ sebagai berikut:<br />

8πhc<br />

E( λ)<br />

5 /<br />

λ<br />

1<br />

=<br />

k T<br />

e hc λ B<br />

Misalkan x=hc/λk B T, maka<br />

E(<br />

λ)<br />

8πk<br />

5 5<br />

B<br />

=<br />

4 4 x<br />

c<br />

h<br />

T<br />

e<br />

x<br />

5<br />

−1<br />

−1<br />

Untuk memperoleh E(λ) maksimum, harus dipenuhi dE/dx=0; jadi,<br />

−<br />

e x<br />

+<br />

1 x −1<br />

=<br />

5<br />

0<br />

x=4,9651<br />

λT=hc/(4,9651 kB)=2,8978x10 -3 mK.<br />

hukum pergeseran Wien<br />

6

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