3. Which of the following represents the correct kinetic energy that an electron would have in the scenario of nuclear beta decay, where electrons are observed to be ejected from the atomic nucleus? Suppose we assume that electrons are somehow trapped within the nucleus, occasionally escaping and being observed in the laboratory. With a typical nucleus diameter of m, use the uncertainty principle to estimate the range of kinetic energies.
(A) 11 MeV
(B) 15 MeV
(C) 19 MeV
(D) 23 MeV
(E) 30 MeV

1. For a muon with a half lifetime and a mass of 250 , (me: mass of an electron) moving with a speed of v = 0.8 c, calculate the mean lifetime τ0, the traveling distance d measured at rest, during τ0, the relativistic momentum P by the unit of eV/c, the de-Broglie wavelength , the phase velocity , the total energy E and kinetic energy (KE) in the unit of eV. (A) , d = 1519 m, P = 102.2 MeV /c, , KE = 35.86 MeV  (B) , d = 1798 m, P = 102.2 MeV/c, , KE = 40.887 MeV  (C) d = 1599 m, P = 113.58 MeV /c, , KE = 43.18 Mev  (D) d =1599 m, , KE = 85.18 Mev  (E) , d =1798 m, , ==3.7475x108m/s

2. For an electron and a neutron with two conditions of kinetic energies: KE=1.0 MeV and 1.0 GeV (M=106, G=109), calculate the momentum P, de-Broglie wavelength , the particle velocity , and phase velocity   (A) For an electron with KE= 1.0 MeV, P = 8.426 MeV Ic, , = 1.5155x108m/s  (B) For an electron with KE= 1.0 GeV, P = 266.44 MeV Ic, , = 4.7925×106m/s  (C) For a neutron with KE= 1.0 GeV, P = 11.43 GeV / c, , = 2.0553 x108m/s  (D) For a neutron with KE= 1.0 MeV, P = 361.34 MeV Ic, , v=1.3829×107 mls  (E) For a neutron with KE=1.0 GeV, P =1.697 GeV I c, , v=2.6226x108m/s  

4. In hydrogen electrons in 2s and 2p state have the same......(choose the right answer) (A) Total energy (B) Average radius (C) Angular momentum (D) More than one of the above is the same (E) None of the above is the same

5. The Heisenberg Uncertainty Principle is generally applied to very small objects such as electrons and protons. Which of the following statements best explains why don't we use uncertainty principle on larger objects such as cars and tennis balls? (A) The errors of a measurement can always, in principle, be made smaller by using more sensitive equipment. (B) Large objects at any instant of time have an exact position and exact momentum and with sufficient care we can measure both precisely. (C) Large objects obey Newton's laws of motion, to which the uncertainty principle does not apply. (D) Because it does apply to large objects, but the uncertainties are so small that we don't notice them. (E) None of above is correct.

6.Rank the velocity of the particles, whose de Broglie waves shown in the figure below ("A" amplitude, and wavelength of the de Broglie waves. All three particles have the same mass.) (A) II ＞I＞ III (B) II ＞I = III (C) III ＞ II=I (D) III ＞ II＞I (E) II =I＞ III

7. Consider the states of a negatively charged hydrogen ion . Which of the following states violates the Pauli exclusion principle? (A)2 electrons both occupying the 1s orbital and carrying opposite spins (B) 1 electron occupying 1s and the other 2s and both carrying the same spin (C) 1 electron occupying 1s and the other 2s and both carrying opposite spins (D) 2 electrons both occupying 2px and carrying the same spin (E) None of the above

8. Consider two spineless, non-interacting particles with identical mass confined in an infinite quantum well. Let E0 and E1 denote ground and first excited states, respectively, of a single particle in the well. Which of the following states is not the correct lowest energy state of the two particles? (A) Both particles being in E0 in the case of identical classical particles; (B) One particle being in E0 and the other in E1 in the case of distinct Fermi particles; (C) Both particles being in E0 in the case of identical Bose particles; (D) Both particles being in E0 in the case of one Fermi and one Bose particles; (E) None of the above.

複選題9.Initially, there are two pairs of electron and positron moving under two conditions: one pair of electronpositron move at the same velocity of 0.6c, side-by-side in the same direction, the other pair of electron and positron move toward each other in an opposite direction with the velocity of -0.6c and 0.8c (c : light velocity). Finally, in the same system, each pair of electron and positron merge together, thus annihilate each other, and totally two photons are generated. Calculate the energies momenta and de-Broglie wavelengths of the electron and positron, two photons move in the same or opposite directions? the momenta of two photons, p1 and p₂, the wavelengths λ1 and λ₂, and the energies of two photons   (A) For positron v = 0.8c: = 408.87 KeV / c, ; v= 0.6c:  = 306.65 KeVIc  (B) For electron v = 0.6c: = 638.86 KeV, = 383.32 KeV / c, (C) Two photons move off in the opposite directions p₁ = 894.4 KeV/ c, p₂ = -596.3 KeV /c (D) Two photons move off p₁ = 1022 KeV/ c, p₂ = -255.5 KeV/ c; = 1022 KeV, = 255.5 KeV (E) Two photons move off in the opposite directions p₁ = 1.9166 MeV /c, p₂ =-851.8 KeV/c,

複選題10. For a neutron with kinetic energy, KE=188 KeV, calculate the velocity V and uncertainty of the position △x to measure this neutron? for an electron with the same △x, calculate the minimum energy E, kinetic energy KE, momentum , velocity , and the maximum De-Broglie wavelength .  (A) For a neutron with KE=188 KeV, = 5.998×106m/s, △x ≥ 4.199×m  (B) For an electron with the same △x, KE ≥1.894 MeV, = 2.35Mev / c, (C) For an electron with the same △x, KE ≥ 8.903 MeV, ≥ 9.4 MeV /c, (D) For an electron with the same △x, E ≥ 9.414 MeV, ≥0.9985 с  (E) For an electron with the same △x, △x ≥ 2.0995×m, ≥4.7 MeV/c,

複選題11. Choose the correct statements about photoelectric effect and properties of photons: (A) The Maxwell equation can perfectly explain the photoelectric effect. (B) The photoelectric effect shows that light comes in quantized chunks of energy. (C) The wavelength of a beam of light is increased but the intensity of light remains unchanged. In this situation, there are more incident photons per second but each photon has less energy. (D) The work function of metal can be measured by using the photoelectric effect. (E) If a piece of metal has a work function of 2.755 eV, photoelectrons can be emitted when it is illuminated with light with a wavelength of 500 nm.

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