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Unformatted text preview: Physics 143 Fall Term 2007 HOMEWORK 5
Due: October 15 Problem 1: (BFG Ch.5 Problem 4)
Suppose that a hydrogen atom in the ground state absorbs a photon of wavelength 15nm. Will the atom be ionized? If so, what will be the kinetic energy of the electron when it gets far away from its atom of origin? The energy of the photon is: EP h f
. hc J . / 1.32 10 J 82.6eV which is larger than E electron is then, 13.6eV. Therefore, the atom can be ionized. The kinetic energy of the K EP E 69eV Problem 2: (BFG Ch.5 Problem 8)
The muon, with mass m 209m , acts as a heavy electron. The muon can bind to a proton to form a muonic atom. Calculate the ionization energy of this atom, and calculate the radius of the muonic atom in its ground state. Ignore reducedmass effect. We know that the ionization energy of a hydrogen atom is: 1 EH m c 2 which is proportional to m. And the radius in its ground state is: 1 a m c which is inversely proportional to m. Therefore, the ionization energy of the muonic atom is: m 209 13.6eV 2.84keV E E m And the radius is: 0.53 m a a 2.54 10 m 0.254pm m 209 (The ionization energy is not the ground state energy. They have the same absolute value but different signs.) Problem 3: (BFG Ch.5 Problem 12)
The one-electron Bohr theory also applies to heavier atoms from which all but one electron have been stripped. We have referred to this kind of atom as a generalized hydrogen atom. Consider carbon, for which Z=6, with only one electron present. What is the wavelength of the photon emitted in the transition from n=3 to n=2? Compare this value with the corresponding one for ordinary hydrogen. The energy of a generalized hydrogen atom is given by, Zc 1 m E n 2
P a g e | 13 Z E n 1 1 68eV 2 3 Therefore, the wavelength of the emitted photon is: hc 18nm E From above, we know that the energy is proportional to Z , so the wavelength of the photo emitted from a carbon atom will be 36 times less than that from a hydrogen atom. E 6 E The transition energy in carbon is: Problem 4: (BFG Ch.6 Problem 1, 2)
1. A time-independent wave function has the form x C exp x /2a . Find C such that | x | dx 1. | x | dx |C| exp x /a dx The integral above is a standard Gaussian Integral. (Appendix B.2) exp x /a dx 1
/ 1 a a Normally, we use a positive C value in wave functions. C 2. Consider the normalized wave function of Problem 1. Calculate the expectation values (a) x , b x , c p and d p for this wave function. Give a physical justification of your results for (a) and (c). Generally, A (a) x (b) x The integral (c) p p 1 a / i a / e 1 a i 1 a / x e dx a a
/ x A x dx 0 1 a / x e dx 0 is odd function, so the integral is 0. a 2 y e dy can be found in Appendix B.2 x i xe e x dx 0 dx is odd function, so the integral is 0.
P a g e | 23 (d) p 1 a e e e a The integral x p and e x e e e dx e dx e dx dx can be found in Appendix B.2 0 shows that the particle is equally likely to be found at x<0 and x>0; 0 shows that the particle is equally likely to be moving to the right and to the left. Problem 5: (BFG Ch.6 Problem 9)
A beam of photons in a range of wavelengths 9.0 1.0nm impinges on an electron in an infinite well of width 1.0nm. The electron is in the ground state. To what excited states can it be boosted (i.e., to what range of values of n can the electron get excited)? The maximum energy of the photons in the beam is: c 6.63 10 J s h E 9.0 1.0 The minimum energy of the photons in the beam is: c 6.63 10 J s E h 9.0 1.0 2.99 10 m/s 10 m 2.99 10 m/s 10 m 2.48 10 J 1.98 10 J The energy level of the infinite well is: (BFG P165 6-45) 6.63 10 J s/2 E 2mL 2 9.1 10 kg 10 m E E E E 6.02 10 J ~20; ~18 The electron in the ground state can be boosted to the 18th, 19th and 20th excited state. P a g e | 33 Problem 6: (BFG Ch.6 Problem 20) Assume the u's are infinite well wave functions.
Consider the wave function x, t the range 0, L/2 as a function of time. What is the period of oscillation of the probability. u exp E u exp E . Calculate the probability that the electron is in The probability that the electron is in the range 0, L/2 is L/ P t x, t x, t dx
L/ L/ L/ u exp u u u u E u exp
E E E E E u exp exp
E E E u exp dx E dx u u exp 2u u cos dx recall cos We know that the infinite well wave functions are: u x E and, P cos
L/ L sin L n 2mL
L L/ u x dx u x dx
E L/ L E dx
L u u cos L sin sin L dx We can use Trig Identity to calculate the integral above:
L/ sin 3x 2x sin dx L L 1 2 1 2 L/ cos x L cos 5x dx L 2L 5 P t 4 5 cos t 5 2mL The period of the oscillation of the probability is: 5 T 2mL 2 T 4mL 5 October 15, 2007 P a g e | 43 ...
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