Answer to Question #106559 in Quantum Mechanics for Ravi kumar

Question

fromplanks law deduce the value of frequency corresponding to peak of (E(not)- V) curve at temp 1000k. In what spectral region does this frequency lie.

Answer 1

The low of Plank

$\epsilon(\nu,T)=\frac{2h\nu^3}{c^2}\cdot\frac{1}{e^{\frac{h\nu}{kT}}-1}.$

$\frac{d\epsilon(\nu,T)}{d\nu}=0$

So, we get that

$-\frac{h\nu}{kT}\cdot\frac{e^{\frac{h\nu}{kT}}}{e^{\frac{h\nu}{kT}}-1}+3=0$ $\to$ $(x-3)e^x+3=0$ $(x=\frac{h\nu}{kT})$

From this equation $x = 2.821439372122$

$\nu=\frac{xkT}{h}=\frac{2.821439372122\cdot1.38\cdot10^{-23}\cdot1000}{6.62\cdot10^{-34}}\approx58.8\cdot10^{12}Hz=58.8THz$

Infrared radiation (IR).