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Chapter 6: Problem 13

Planck's radiation law tells us that the intensity of radiation per unit area and per unit wavelength \(\lambda\) from a black body at temperature \(T\) is $$ I(\lambda)=\frac{2 \pi h c^{2} \lambda^{-5}}{\mathrm{e}^{h c / \lambda k_{g} T}-1} $$ where \(h\) is Planck's constant, \(c\) is the speed of light, and \(k_{B}\) is Boltzmann's constant. a) Show by differentiating that the wavelength \(\lambda\) at which the emitted radiation is strongest is the solution of the equation $$ 5 \mathrm{e}^{-h c / \lambda k_{B} T}+\frac{h c}{\lambda k_{B} T}-5=0 . $$ Make the substitution \(x=h c / \lambda k_{B} T\) and hence show that the wavelength of maximum radiation obeys the Wien displacement law: $$ \lambda=\frac{b}{T} $$ where the so-called Wien displacement constant is \(b=h c / k_{B} x\), and \(x\) is the solution to the nonlinear equation $$ 5 e^{-x}+x-5=0 $$ b) Write a program to solve this equation to an accuracy of \(\epsilon=10^{-6}\) using the binary search method, and hence find a value for the displacement constant. c) The displacement law is the basis for the method of optical pyrometry, a method for measuring the temperatures of objects by observing the color of the thermal radiation they emit. The method is commonly used to estimate the surface temperatures of astronomical bodies, such as the Sun. The wavelength peak in the Sun's emitted radiation falls at \(\lambda=502 \mathrm{~nm}\). From the equations above and your value of the displacement constant, estimate the surface temperature of the Sun.

Short Answer

Expert verified
Differentiate Planck's law, set to zero, use substitution \( x= \frac{hc}{\lambda k_B T}\), binary search to solve \(5e^{-x}+x-5=0\), and estimate the Sun's temperature with \( \lambda=502 \mathrm{nm}.\)

Step by step solution

01

Differentiate Planck's radiation law

We start by differentiating the given Planck's radiation law with respect to wavelength \(\lambda\):\(I(\lambda)=\frac{2 \pi h c^{2} \lambda^{-5}}{e^{hc / \lambda k_B T} -1}\). To find the peak intensity, we need to find where the derivative equals zero.
02

Apply chain and product rules

Use the chain rule and product rule to differentiate\( I(\lambda) \):\(\frac{dI}{d\lambda}\). The differentiation requires handling the exponential and the polynomial terms separately.
03

Set the derivative to zero

Set the differentiated expression to zero to find the critical points: \(\frac{dI}{d\lambda} = 0.\)
04

Simplify the resulting equation

You will get an expression involving exponentials and polynomials. Substitute \(x=\frac{hc}{\lambda k_B T}\) into the equation. This leads to the simplified equation for critical points: \(5e^{-x} + x - 5 = 0\).
05

Show Wien's displacement law

Rearrange values to prove Wien’s displacement law: \(\lambda = \frac{hc}{k_B T x} = \frac{b}{T}\). Here, \(b=\frac{hc}{k_B x}\) represents the Wien displacement constant.
06

Solve the nonlinear equation

Use a programming language capable of binary search to solve \(5e^{-x} + x - 5 = 0\) accurately within \(\epsilon=10^{-6}\). Your goal is to find the value of \(x \).
07

Calculate the Wien displacement constant

Using the found value of \( x \), calculate the Wien displacement constant \(b=\frac{hc}{k_B x}.\)
08

Estimate the Sun's surface temperature

Given the peak wavelength of the Sun’s radiation \( \lambda = 502 \mathrm{ nm},\) use Wien’s law to estimate its temperature: \( T = \frac{b}{\lambda}.\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wien's Displacement Law
Wien's Displacement Law is essential in understanding black body radiation.
It states that the wavelength (\( \lambda \)) at which a black body emits radiation most strongly is inversely proportional to its temperature (\( T \)).
Mathematically, this is given by \ \[ \ \lambda = \frac{b}{T} \ \] where \ \(b \) is the Wien displacement constant.
This law helps determine that hotter objects radiate energy at shorter wavelengths compared to cooler objects.
\ \ For example, the blue color of a hotter flame versus the red color of a cooler flame.
\ \ In the given exercise, by substituting \ \(x = \frac{hc}{\lambda k_B T} \) into the differentiated Planck's radiation law, we derive the equation \ \[ 5 \ e^{-x} + x - 5 = 0 \]. Solving this yields the value of \ \(x \), which allows us to find \ \(b \), since \ \( b = \frac{hc}{k_B x} \).
\ \ This forms the basis of optical pyrometry, which measures temperature by observing the emitted radiation color.
Binary Search Method
The Binary Search Method is an efficient algorithm used to find the solution of equations especially when you need precise results.
\ \ In our problem, for solving \ \[ 5 \ e^{-x} + x - 5 = 0 \] within an accuracy of \ \( \epsilon = 10^{-6} \). Binary search helps to narrow down the values iteratively.
\ \ Steps to implement Binary Search Method:
1. Choose an initial range within which the solution lies.
2. Calculate the midpoint of the range. If the function value at this midpoint satisfies the equation within the desired accuracy, it is our solution.
3. If the function value is not zero, determine whether the root lies in the left or right half of the current range and update the range accordingly.
4. Repeat steps 2 and 3 until the precision \ \( \epsilon \) is achieved.
\ \ This method helps us find the exact value of \ \(x \), which is necessary for calculating the Wien displacement constant.
Differentiation in Physics
Differentiation is a crucial tool in physics for understanding how a quantity changes with respect to another.
\ \ In the context of Planck's radiation law, differentiation helps us find the point at which the intensity of radiation is at its peak.
By differentiating the intensity function \ \[ I(\lambda) = \frac{2\pi hc^2 \lambda^{-5}}{e^{hc/\lambda k_B T} - 1} \], we can determine its critical points.
\ \ Using the chain and product rules, we handle the exponential and polynomial terms, setting the resulting derivative to zero.
\ \ The critical points where the derivative is zero indicate the wavelengths at which the radiation intensity is strongest.
\ \ This application illustrates the power of differentiation in finding maximum or minimum values, which is fundamental in many areas of physics.

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Most popular questions from this chapter

voltages \(V_{1} \ldots V_{N}\) satisfy the equations $$ \begin{aligned} 3 V_{1}-V_{2}-V_{3} &=V_{+,} \\ -V_{1}+4 V_{2}-V_{3}-V_{4} &=V_{+,} \\ \vdots & \\ -V_{i-2}-V_{i-1}+4 V_{i}-V_{i+1}-V_{i+2} &=0, \\ \vdots V_{N-3}-V_{N-2}+4 V_{N-1}-V_{N} &=0, \\ -V_{N-2}-V_{N-1}+3 V_{N} &=0 . \end{aligned} $$ Express these equations in vector form \(\mathbf{A v}=\mathbf{w}\) and find the values of the matrix \(\mathrm{A}\) and the vector \(\mathrm{w}\). b) Write a program to solve for the values of the \(V_{i}\) when there are \(N=6\) internal junctions with unknown voltages. (Hint: All the values of \(V_{i}\) should lie between zero and 5V. If they don't, something is wrong.) c) Now repeat your calculation for the case where there are \(N=10000\) internal junctions. This part is not possible using standard tools like the solve function. You need to make use of the fact that the matrix \(\mathrm{A}\) is banded and use the banded function from the file banded.py, discussed in Appendix E.

A chain of resistors Consider a long chain of resistors wired up like this: All the resistors have the same resistance \(R\). The power rail at the top is at voltage \(V_{+}=\) \(5 V\). The problem is to find the voltages \(V_{1} \ldots V_{N}\) at the internal points in the circuit. a) Using Ohm's law and the Kirchhoff current law, which says that the total net current flow out of (or into) any junction in a circuit must be zero, show that the voltages \(V_{1} \ldots V_{N}\) satisfy the equations $$ \begin{aligned} 3 V_{1}-V_{2}-V_{3} &=V_{+,} \\ -V_{1}+4 V_{2}-V_{3}-V_{4} &=V_{+,} \\ \vdots & \\ -V_{i-2}-V_{i-1}+4 V_{i}-V_{i+1}-V_{i+2} &=0, \\ \vdots V_{N-3}-V_{N-2}+4 V_{N-1}-V_{N} &=0, \\ -V_{N-2}-V_{N-1}+3 V_{N} &=0 . \end{aligned} $$ Express these equations in vector form \(\mathbf{A v}=\mathbf{w}\) and find the values of the matrix \(\mathrm{A}\) and the vector \(\mathrm{w}\). b) Write a program to solve for the values of the \(V_{i}\) when there are \(N=6\) internal junctions with unknown voltages. (Hint: All the values of \(V_{i}\) should lie between zero and 5V. If they don't, something is wrong.) c) Now repeat your calculation for the case where there are \(N=10000\) internal junctions. This part is not possible using standard tools like the solve function. You need to make use of the fact that the matrix \(\mathrm{A}\) is banded and use the banded function from the file banded.py, discussed in Appendix E.

There is a magical point between the Earth and the Moon, called the \(L_{1}\) Lagrange point, at which a satellite will orbit the Earth in perfect synchrony with the Moon, staying always in between the two. This works because the inward pull of the Earth and the outward pull of the Moon combine to create exactly the needed centripetal force that keeps the satellite in its orbit. Here's the setup: a) Assuming circular orbits, and assuming that the Earth is much more massive than either the Moon or the satellite, show that the distance \(r\) from the center of the Earth to the \(L_{1}\) point satisfies $$ \frac{G M}{r^{2}}-\frac{G m}{(R-r)^{2}}=\omega^{2} r $$ where \(R\) is the distance from the Earth to the Moon, \(M\) and \(m\) are the Earth and Moon masses, \(G\) is Newton's gravitational constant, and \(\omega\) is the angular velocity of both the Moon and the satellite. b) The equation above is a degree-five polynomial equation in \(r\) (also called a quintic equation). Such equations cannot be solved exactly in closed form, but it's straightforward to solve them numerically. Write a program that uses either Newton's method or the secant method to solve for the distance \(r\) from the Earth to the \(L_{1}\) point. Compute a solution accurate to at least four significant figures. The values of the various parameters are: $$ \begin{aligned} G &=6.674 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2} \\\ M &=5.974 \times 10^{24} \mathrm{~kg} \\ m &=7.348 \times 10^{22} \mathrm{~kg} \\ R &=3.844 \times 10^{8} \mathrm{~m} \\ \omega &=2.662 \times 10^{-6} \mathrm{~s}^{-1} \end{aligned} $$ You will also need to choose a suitable starting value for \(r\), or two starting values if you use the secant method.

The biochemical process of glycolysis, the breakdown of glucose in the body to release energy, can be modeled by the equations $$ \frac{\mathrm{d} x}{\mathrm{~d} t}=-x+n y+x^{2} y_{0} \quad \frac{\mathrm{d} y}{\mathrm{~d} t}=b-a y-x^{2} y $$ Here \(x\) and \(y\) represent concentrations of two chemicals, ADP and F6P, and \(a\) and \(b\) are positive constants. One of the important features of nonlinear equations like these is their stationary points, meaning values of \(x\) and \(y\) at which the derivatives of both variables become zero simultaneously, so that the variables stop changing and become constant in time. Setting the derivatives to zero above, the stationary points of our glycolysis equations are solutions of $$ -x+a y+x^{2} y=0, \quad b-a y-x^{2} y=0 $$ a) Demonstrate analytically that the solution of these equations is $$ x=b, \quad y=\frac{b}{a+b^{2}} $$ b) Show that the equations can be rearranged to read $$ x=y\left(a+x^{2}\right), \quad y=\frac{b}{a+x^{2}} $$ and write a program to solve these for the stationary point using the relaxation method with \(a=1\) and \(b=2\). You should find that the method fails to converge to a solution in this case. c) Find a different way to rearrange the equations such that when you apply the relaxation method again it now converges to a fixed point and gives a solution. Verify that the solution you get agrees with part (a).

Consider the degree-six polynomial $$ P(x)=924 x^{6}-2772 x^{5}+3150 x^{4}-1680 x^{3}+420 x^{2}-42 x+1 $$ There is no general formula for the roots of a polynomial of degree six, but one can find them easily enough using a computer. a) Make a plot of \(P(x)\) from \(x=0\) to \(x=1\) and by inspecting it find rough values for the six roots of the polynomial-the points at which the function is zero. b) Write a Python program to solve for the positions of all six roots to at least ten decimal places of accuracy, using Newton's method. Note that the polynomial in this example is just the sixth Legendre polynomial (mapped onto the interval from zero to one), so the calculation performed here is the same as finding the integration points for 6-point Gaussian quadrature (see Section 5.6.2), and indeed Newton's method is the method of choice for calculating Gaussian quadrature points.
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