Particles Behaving as Waves Problems and Solutions 3

Q#39.61

The Red Supergiant Betelgeuse. The star Betelgeuse has a surface temperature of 3000 K and is 600 times the diameter of our sun. (If our sun were that large, we would be inside it!) Assume that it radiates like an ideal blackbody. 

(a) If Betelgeuse were to radiate all of its energy at the peak-intensity wavelength, how many photons per second would it radiate? 

(b) Find the ratio of the power radiated by Betelgeuse to the power radiated by our sun (at 5800 K). 

Answer:
Assuming that Betelgeuse radiates like a perfect blackbody, Wien’s displacement and the Stefan-Boltzmann law apply to its radiation. 

Wien’s displacement law is

$\lambda_{peak}=\frac{2.90 \times 10^{-3} \ m.K}{T}$

and the Stefan-Boltzmann law says that the intensity of the radiation is $I = \sigma T^4$

so the total radiated power is $P = \sigma AT^4$

$A = 4\pi r^2 = 4\pi (600 \times 6.96 \times 10^8 \ m)^2 = 2.19 \times 10^{24} \ m$

(a) First use Wien’s law to find the peak wavelength: 

$\lambda_{m}=\frac{2.90 \times 10^{-3} \ m.K}{3000 \ K}=9.667 \times 10^{-7} \ m$

Call N the number of photons/second radiated

N x (energi per photon) = IA = $\sigma AT^4$

$N \frac{hc}{\lambda_m}= \sigma AT^4$

$N = \frac{\lambda_m \sigma AT^4}{hc}$

$N = \frac{(9.667 \times 10^{-7} \ m)(5.67 \times 10^{-8} \ W.m^{-2}.K^{-4})(2.19 \times 10^{24} \ m^2)(3000 \ K)^4}{(6.626 \times 10^{-34} \ J.s)(3.00 \times 10^8 \ m/s)}$

$N = 5 \times 10^{49}$ photon/s

(b) $\frac{I_B A_B}{I_S A_S}=\frac{\sigma A_B T_B^4}{\sigma A_S T_S^4}$

 $\frac{I_B A_B}{I_S A_S}=\frac{4\pi R_B^2 T_B^4}{4 \pi R_S^2 T_S^4}$

 $\frac{I_B A_B}{I_S A_S}=\frac{R_B^2 T_B^4}{R_S^2 T_S^4}$

 $\frac{I_B A_B}{I_S A_S}=\left(\frac{600 R_S}{R_S}\right)^2 \left(\frac{300 \ K}{5800 \ K}\right)^4 = 3 \times 10^4$

Betelgeuse radiates 30,000 times as much energy per second as does our sun!  

Q#39.62

CP Light from an ideal spherical blackbody 15.0 cm in diameter is analyzed using a diffraction grating having When you shine this light through the grating, you observe that the peak-intensity wavelength forms a first-order bright fringe at $\pm 11.6^0$from the central bright fringe. (a) What is the temperature of the blackbody? (b) How long will it take this sphere to radiate 12.0 MJ of energy?  

Answer:
The diffraction grating allows us to determine the peak-intensity wavelength of the light. Then Wien’s displacement law allows us to calculate the temperature of the blackbody, and the StefanBoltzmann law allows us to calculate the rate at which it radiates energy.

The bright spots for a diffraction grating occur when d sin θ = mλ. Wien’s displacement law is 

$\lambda_{peak}=\frac{2.90 \times 10^{-3} \ m.K}{T}$

and the Stefan-Boltzmann law says that the intensity of the radiation is $I = \sigma T^4$

so the total radiated power is $P = \sigma AT^4$

(a) First find the wavelength of the light:

λ = d sin θ = $\frac{1}{385,000 \ lines/m} sin \ 11.6^0 = 5.22 \times 10^{-7} \ m$

Now use Wien’s law to find the temperature:

$T=\frac{2.90 \times 10^{-3} \ m.K}{\lambda}$

$T=\frac{2.90 \times 10^{-3} \ m.K}{5.22 \times 10^{-7} \ m}=5550 \ K$

(b) The energy radiated by the blackbody is equal to the power times the time, giving

$U= Pt= IAt = \sigma AT^4 t$  which gives

$t = \frac{U}{\sigma AT^4}$

$t = \frac{12.0 \times 10^6 \ J}{(5,67 \times 10^8 \ W.m^{-2}.K^{-4})(4 \pi)(0.0750 \ m)^2(5550 \ K)^4}=3.16 \ s$

By ordinary standards, this blackbody is very hot, so it does not take long to radiate 12.0 MJ of energy. 

 

Q#39.63

What must be the temperature of an ideal blackbody so that photons of its radiated light having the peak-intensity wavelength can excite the electron in the Bohr-model hydrogen atom from the ground state to the third excited state?  

Answer:

The energy of the peak-intensity photons must be equal to the energy difference between the n =1 and the n = 4 states. Wien’s law allows us to calculate what the temperature of the blackbody must be for it to radiate with its peak intensity at this wavelength.

In the Bohr model, the energy of an electron in shell n is

$E_n=\frac{-13.6 \ eV}{n^2}$ 

and Wien’s displacement law is

$\lambda_{m}=\frac{2.90 \times 10^{-3} \ m.K}{T}$

The energy of a photon is $E = hf=\frac{hc}{\lambda}$

First find the energy (ΔE) that a photon would need to excite the atom. The ground state of the atom is  n = 1 and the third excited state is n = 4. This energy is the difference between the two energy levels. Therefore 

$\Delta E = (-13.6 \ eV)\left(\frac{1}{4^2}-\frac{1}{1^2}\right)=12.8 \ eV$

Now find the wavelength of the photon having this amount of energy.

$\frac{hc}{\lambda}=12.8 \ eV$

$\lambda=\frac{hc}{12.8 \ eV}$

$\lambda=\frac{(4.136 \times 10^{-15} eV.s)(3.00 \times 10^8 \ m/s)}{12.8 \ eV}$

$\lambda=9.73 \times 10^{-8} \ m$

Now use Wien’s law to find the temperature.

$T=\frac{2.90 \times 10^{-3} \ m.K}{\lambda}$

$T=\frac{2.90 \times 10^{-3} \ m.K}{9.73 \times 10^{-8} \ m}=29800 \ K$

This temperature is well above ordinary room temperatures, which is why hydrogen atoms are not in excited states during everyday conditions. 

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