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1· Plot graphs using each expression at different frequencies and different temperatures (1000 K, 2000 K, 3000 K, 4000 K, 5000 K, 6000 K). Compare and contrast the graphs you get on each function Equation 1 8rhv3 1 dv Equation 2 Use these sample frequencies for your plots Frequency Frequency Frequency 2.99E+12 3.7375E+12 9.97E+13 1.50E+14 1.50E+1 1.99E+15 4.98333E+12 1.99E+14 2.99E+15 2.99E+14 7.475E+12 9.97E+12 1.50E+13 2.99E+13 3.74E+13 4.98E+13 5.98E+13 7.48E+13 3.32E+14 3.74E+14 4.27E+14 4.98E+14 5.98E+14 7.48E+14 9.97E+1 What are the names associated with each expression? Equation 1: 2. Equation 2: 3. Briefly explain the effect of temperature and frequency on blackbody radiation IF 4. Represent Equation 2 above as a function of wavelength Q)
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Answer #1

i have used gnuplot to plot the function rho for different temperature

] Gnuplot (window id : 0) 4x106 f(x) 3.5x106 6 3x10 2.5x106 2x106 1.5x106 1x106 500000 5x10% 1x10-6 1.5x106 2x10-6 2.5x10-6 3this is distribution graph for T = 6000 K

and coding in gnuplot is :

gnuplot File Plot Expressions Functions General Axes Chart Styles 3D Help Terminal type set to wxt gnuplot> pi 3.14 gnuplotgnuplot File Plot Expressions Functions General Axes Chart Styles 3D Help Enuplot> plot f(x) undefined variable: exp gnuplot> f(x)(nr)((x*5) (exp( (e_x/x)-1.0))) Enuplot> plot f(x) gnuplot> set xrange [10++-6 : 3+1044-6] Enuplot> plot f(x) gnuplot> set xrange [-1**-6 3 184-6] Enuplot> plot f(x) gnuplot> set xrange [-1.8 1-631844-6] Enuplot> plot f(x) gnuplot> set xrange e 18*-63 1**-6] Enuplot> plot f(x) gnuplot> T75e Enuplot> (hc)/(k*T) gnuplot> f(x)(nr((x*5) (exp( (e x/x)-1.0))) Enuplot> set xrange [e 1e*-63 10**-6 gnuplot> plot f(x) gnuplot> T-1-6800 , θ gnuplot T 2 750.e Enuplot> T3300.0 gnuplot> T-4 = 88.0 undefined variable: T1 gnuplot> ex(hc)(k T1) gnuplot> ex3(h*c)/(kT 3) gnuplot> f(x)(nr((x*5 (exp( (e x1/x)-1.0))) Enuplot> f20x)nr)(x**S)*(exp((ex2/x)-1.0))) gnuplot> f3(x)(nr)((x**5) (exp((e x3/x)-1.0))) Enuplot f4(x) -(n_r)J((x**5) (exp (x4/x)-1.0))) gnuplot > set xrange [o-10..-5 : 37 .0+10++-6) Enuplot> plot f4(x) gnuplot> set xrange e 18*-639.*1844-6] Enuplot> plot f4(x) gnuplot> set xrange e 18*-6 49.9*1844-6] Enuplot> plot f4(x) gnuplot> set xrange e 18*-689.*1844-6] Enuplot> plot f4(x) gnuplot> set xrange e 18*-6 99.9*1844-6] Enuplot> plot f4(x) gnuplot> plot f4(x) f3(x) Enuplot> 11:54 AM O Type here to search 1/16/2019

we are getting a peak at lambda = 0.5 micro meter approximately from graph and theoretically it is verified from wein's law as lambda max theoretically is = 0.4833 micro meter 483.3 nm = blue color

Wein lauw 0 009 Km max 〉、max. 3866-7 nm May Max

equation 1 is Classical physics can be used to derive an equation which describes the intensity of blackbody radiation as a function of frequency for a fixed temperature — the result is known as the Rayleigh-Jeans law

equation 2 is planck's law

The Rayleigh-Jeans Law was an important step in our understanding of the equilibrium radiation from a hot object, even though it turned out not to be an accurate description of nature. The careful work in developing the Rayleigh-Jeans law laid the foundation for the quantum understanding expressed in the Planck radiation formula.

in terms of wavelength expression is

ho(lambda ,T) = dfrac{8 pi h c}{lambda^5}dfrac{1}{e^{hc/lambda kT}-1}

please rate it up thanks :)

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