Q#39.58
A large number of hydrogen atoms are in thermal equilibrium. Let be the ratio $\frac{n_2}{n_1}$ of the number of atoms in an excited state to the number of atoms in an ground state. At what temperature is $\frac{n_2}{n_1}$ equal to
(a) $10^{-12}$;
(b) $10^{-8}$,
(c) $10^{-4}$
(d) Like the sun, other stars have continuous spectra with dark absorption lines (see Fig. 39.9). The absorption takes place in the star’s atmosphere, which in all stars is composed primarily of hydrogen. Explain why the Balmer absorption lines are relatively weak in stars with low atmospheric temperatures such as the sun (atmosphere temperature 5800 K) but strong in stars with higher atmospheric temperatures.
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| Fig.39.9 |
We use $\frac{n_2}{n_2}=e^{-\frac{(E_x-E_g)}{kT}}$
$ln\frac{n_2}{n_2}=-\frac{(E_x-E_g)}{kT}$
$T = \frac{-(E_x-E_g)}{kln\frac{n_2}{n_1}}$
with $E_x=E_2=\frac{-13.6 \ eV}{2^2}=-3.4 \ eV$ and $E_g=-13.6 \ eV$, so
$E_x-E_g=-3.4 \ eV - (-13.6 \ eV)=10.2 \ eV = 1.63 \times 10^{-18} \ J$
(a) so, for $\frac{n_2}{n_1}=10^{-12}$;
$T = \frac{-(1.63 \times 10^{-18} \ J)}{(1.38 \times 10^{-23} \ J/K)ln(10^{-12})} = 4275 \ K$
(b) so, for $\frac{n_2}{n_1}=10^{-8}$;
$T = \frac{-(1.63 \times 10^{-18} \ J)}{(1.38 \times 10^{-23} \ J/K)ln(10^{-8})} = 6412 \ K$
(c) so, for $\frac{n_2}{n_1}=10^{-4}$;
$T = \frac{-(1.63 \times 10^{-18} \ J)}{(1.38 \times 10^{-23} \ J/K)ln(10^{-4})} = 12824 \ K$
Q#39.59
A sample of hydrogen atoms is irradiated with light with wavelength 85.5 nm, and electrons are observed leaving the gas.
(a) If each hydrogen atom were initially in its ground level, what would be the maximum kinetic energy in electron volts of these photoelectrons?
(b) A few electrons are detected with ener[1]gies as much as 10.2 eV greater than the maximum kinetic energy calculated in part (a). How can this be?
Answer:
(a) The photon energy is given to the electron in the atom. Some of this energy
overcomes the binding energy of the atom and what is left appears as kinetic energy of the free electron.
Apply $hf=E_f - E_i$, the energy given to the electron in the atom when a photon is absorbed.
The energy of one photon is
$\frac{hc}{\lambda}=\frac{(6.626 \times 10^{-34} \ J.s)(2.998 \times 10^8 \ m/s)}{85.5 \times 10^{-9} \ m}$
$\frac{hc}{\lambda}=2.323 \times 10^{-18} \ J$
$\frac{hc}{\lambda}=\frac{2.323 \times 10^{-18} \ J}{1.60 \times 10^{-19} \ J/eV} = 14.50 \ eV$
The final energy of the electron is $E_f= hf+E_i$. In the ground state of the hydrogen atom the energy of the electron is $E_i=-13.6 \ eV$. Thus
$E_f=-13.6 \ eV + 14.50 \ eV = 0.90 \ eV$
(b) At thermal equilibrium a few atoms will be in the 2 n = excited levels, which have an energy of $\frac{-13.6 \ eV}{4}=-3.4 \ eV$, 10.2 eV greater than the energy of the ground state.
If an electron with E = − 3.40 eV gains 14.5 eV from the absorbed photon, it will end up with
14 5 eV − 3.4 eV = 11.1 eV of kinetic energy.
Q#39.60
CP Bohr Orbits of a Satellite. A 20.0-kg satellite circles the earth once every 2.00 h in an orbit having a radius of 8060 km.
(a) Assuming that Bohr’s angular-momentum result ($nh/2\pi$) applies to satellites just as it does to an electron in the hydrogen atom, find the quantum number n of the orbit of the satellite.
(b) Show from Bohr’s angular momentum result and Newton’s law of gravitation that the radius of an earth-satellite orbit is directly proportional to the square of the quantum number $r=kn^2$, where k is the constant of proportionality.
(c) Using the result from part (b), find the distance between the orbit of the satel[1]lite in this problem and its next “allowed” orbit. (Calculate a numerical value.)
(d) Comment on the possibility of observing the separation of the two adjacent orbits.
(e) Do quantized and classi[1]cal orbits correspond for this satellite? Which is the “correct” method for calculating the orbits?
Answer:
For circular motion, L = mvr, and $a = \frac{v^2}{r}$
Newton’s law of gravitation is $F_g=G\frac{Mm}{r^2}$ with $G = 6.67 \times 10{-11} \ N.m^2/kg^2$
The period T is 2.00 h 7200 s
(a) F = ma
$G\frac{m.m_E}{r^2}=m\frac{v^2}{r}$
$v^2=G\frac{m_E}{r}$
The Bohr postulate says
$v=\frac{nh}{2\pi mr}$, so
$\left(\frac{nh}{2\pi mr}\right)^2=G\frac{m_E}{r}$
$\left(\frac{nh}{2\pi m}\right)^2r=Gm_E$
$r=\left(\frac{h^2}{4\pi^2 Gm_E m^2}\right)n^2$
This is in the form $r=kn^2$ with
$k = \frac{h^2}{4\pi^2 Gm_E m^2}$
$k = \frac{(6.626 \times 10^{-34} \ J.s)^2}{4\pi^2 (6.67 \times 10^{-11} \ N.m^2/kg^2)(5.97 \times 10^{24} \ kg) (20 \ kg)^2}$
$k = 7.0 \times 10^{-86} \ m$
(c) $\Delta r = r_{n+1}-r_n=k([n+1]^2-n^2)=(2n+1)k$
$\Delta r = 2(1.08 \times 10^{46})(7.0 \times ^{-86} \ m) = 1.5 \times 10^{-39} \ m$
(d) Δr is exceedingly small, so the separation of adjacent orbits is not observable.
(e) There is no measurable difference between quantized and classical orbits for this satellite; either method of calculation is totally acceptable.
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