Particles Behaving as Waves Problems and Solutions 4

Q#39.64

CP An ideal spherical blackbody 24.0 cm in diameter is maintained at 225 $^0C$ by an internal electrical heater and is immersed in a very large open-faced tank of water that is kept boiling by the energy radiated by the sphere. You can neglect any heat transferred by conduction and convection. Consult Table 17.4 as needed. 

(a) At what rate, in g/s is water evaporating from the tank? 

(b) If a physics-wise thermophile organism living in the hot water is observing this process, what will it measure for the peakintensity (i) wavelength and (ii) frequency of the electromagnetic waves emitted by the sphere? 

Answer:

The blackbody radiates heat into the water, but the water also radiates heat back into the blackbody. The net heat entering the water causes evaporation. Wien’s law tells us the peak wavelength radiated, but a thermophile in the water measures the wavelength and frequency of the light in the water.

By the Stefan-Boltzman law, the net power radiated by the blackbody is 

$\frac{dQ}{dt}=\sigma A(T_{sphere}^4-T_{water}^4)$

Since this heat evaporates water, the rate at which water evaporates is

$\frac{dQ}{dt}=L_v\frac{dm}{dt}$

Wien’s displacement law is

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

 and the wavelength in the water is $\lambda_w=\frac{\lambda_0}{n}$

(a) The net radiated heat is 

$\frac{dQ}{dt}=\sigma A(T_{sphere}^4-T_{water}^4)$

 and the evaporation rate is  $\frac{dQ}{dt}=L_v\frac{dm}{dt}$

 where dm is the mass of water that evaporates in time dt. Equating these two rates gives

$L_v\frac{dm}{dt}=\sigma A(T_{sphere}^4-T_{water}^4)$

$\frac{dm}{dt}=\frac{\sigma 4\pi R^2(T_{sphere}^4-T_{water}^4)}{L_v}$

$\frac{dm}{dt}=\frac{(5.67 \times 10^{-8} W.m^{-2}.K^{-4})(4\pi)(0.120 \ m)^2[(498 \ K)^4-(373 \ K)^4)}{2256 \times 10^3 \ J/kg}$

$\frac{dm}{dt}=1.92 \times 10^{-4} \ kg/s = 0.192 \ g/s$

(b) (i) Wien’s law gives $\lambda_{m}=\frac{2.90 \times 10^{-3} \ m.K}{498 \ K}=5.82 \times 10^{-6} \ m$.

But this would be the wavelength in vacuum. In the water the thermophile organism would measure

$\lambda_w=\frac{\lambda_0}{n} = \frac{5.82 \times 10^{-6} \ m}{1.333}= 4.37 \times 10^{-6} \ m = 4.37 \mu m$

(ii) The frequency is the same as if the wave were in air, so

$f = \frac{c}{\lambda_0}$

$f=\frac{3.00 \times 10^8 \ m/s}{5.82 \times 10^{-6} \ m}=5.15 \times 10^{13} \ Hz$ 

Q#39.65 

When a photon is emitted by an atom, the atom must recoil to conserve momentum. This means that the photon and the recoiling atom share the transition energy. 

(a) For an atom with mass m, calculate the correction $\Delta \lambda$due to recoil to the wavelength of an emitted photon. Let $\lambda$ be the wavelength of the photon if recoil is not taken into consideration. (Hint: The correction is very small, as Problem 39.56 suggests, so $|\Delta \lambda|/\lambda  << 1$. Use this fact to obtain an approximate but very accurate expression for $\lambda$.) 

(b) Evaluate the correction for a hydrogen atom in which an electron in the nth level returns to the ground level. How does the answer depend on n?

Answer:

Apply conservation of energy and conservation of linear momentum to the system of atom plus photon. 

Let $E_{tr}$ be the transition energy, $E_{ph}$ be the energy of the photon with wavelength λ′, and $E_{r}$ be the kinetic energy of the recoiling atom. Conservation of energy gives

$E_{ph}+E_r=E_{tr}$

with $E_{ph}=\frac{hc}{\lambda'}$, so

$\frac{hc}{\lambda'}=E_{tr}-E_r$

$\lambda'=\frac{hc}{E_{tr}-E_r}$

If the recoil energy is neglected then the photon wavelength is $\lambda=\frac{hc}{E_{tr}}$

$\Delta \lambda =\lambda '-\lambda $

$\Delta \lambda =\left ( \frac{hc}{E_{tr}-E_r}-\frac{hc}{E_{tr}} \right )$

$\Delta \lambda =\left ( \frac{hc}{E_{tr}} \right )\left ( \frac{1}{1-E_r/E_{tr}}-1 \right )$

with $\frac{1}{1-E_r/E_{tr}}=\left ( 1-\frac{E_r}{E_{tr}} \right )^{-1}\approx 1+\frac{E_r}{E_{tr}}$ since $\frac{E_r}{E_{tr}}<<1$

Thus, $\Delta \lambda =\left ( \frac{hc}{E_{tr}} \right )\left ( \frac{E_r}{E_{tr}} \right )$

or since $E_{tr}=\frac{hc}{\lambda}$

$\Delta \lambda =\left ( \frac{E_r}{hc} \right )\lambda^2$

(b) Use conservation of linear momentum to find $E_r$ : Assuming that the atom is initially at rest, the momentum $p_r$ of the recoiling atom must be equal in magnitude and opposite in direction to the momentum $p_{ph}=\frac{h}{\lambda}$ of the emitted photon: $\frac{h}{\lambda}=p_{r}$

$E_r=\frac{p_r^2}{2m}$

where m is the mass of the atom, so

$E_r=\frac{h^2}{2m\lambda^2}$

Use this result in the above equation: 

$\Delta \lambda =\left ( \frac{E_r}{hc} \right )\lambda^2=\left ( \frac{\frac{h^2}{2m\lambda^2}}{hc} \right )\lambda^2$

$\Delta \lambda =\frac{h}{2mc}$

note that this result for Δλ is independent of the atomic transition energy.

(b) For a hydrogen atom m = $m_p$ and   

$\Delta \lambda =\frac{h}{2m_p c}$

$\Delta \lambda =\frac{6.626 \times 10^{-34} \ J.s}{2(1.673 \times 10^{-27} \ kg)(2.998 \times 10^8) \ m/s}=6.61 \times 10^{-16} \ m$

Q#39.66

An Ideal Blackbody. A large cavity with a very small hole and maintained at a temperature T is a good approximation to an ideal radiator or blackbody. Radiation can pass into or out of the cavity only through the hole. The cavity is a perfect absorber, since any radiation incident on the hole becomes trapped inside the cavity. Such a cavity at 200°C has a hole with area 4.00 $mm^2$. How long does it take for the cavity to radiate 100 J of energy through the hole?

Answer:

Combine $I = \sigma T^4$, P = IA and $\Delta E = Pt$.

In the Stefan-Boltzmann law the temperature must be in kelvins. 200 $^0$C = 473 K

so, $t = \frac{\Delta E}{A\sigma T^4}$

$t = \frac{100 \ J}{(4.00 \times 10^{-6} \ m^2)(5.67 \times 10^{-8} \ W.m^{-2}.K^{-4})(473 \ K)^4}$

$t = 8.81 \times 10^3 \ s= 2.45 \ h$

P = 0.0114 W. Since the area of the hole is small, the rate at which the cavity radiates energy through the hole is very small. 

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