Question #106559
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.
1
Expert's answer
2020-03-26T12:16:43-0400

The low of Plank


ϵ(ν,T)=2hν3c21ehνkT1.\epsilon(\nu,T)=\frac{2h\nu^3}{c^2}\cdot\frac{1}{e^{\frac{h\nu}{kT}}-1}.


dϵ(ν,T)dν=0\frac{d\epsilon(\nu,T)}{d\nu}=0


So, we get that


hνkTehνkTehνkT1+3=0-\frac{h\nu}{kT}\cdot\frac{e^{\frac{h\nu}{kT}}}{e^{\frac{h\nu}{kT}}-1}+3=0 \to (x3)ex+3=0(x-3)e^x+3=0 (x=hνkT)(x=\frac{h\nu}{kT})


From this equation x=2.821439372122x = 2.821439372122


ν=xkTh=2.8214393721221.38102310006.62103458.81012Hz=58.8THz\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).


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