Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 43 | a 24 Si has 14 protons and 14 neutrons. b 37Rb has 37 protons and 48 neutrons. c 205Tl has 81 protons and 124 neutrons. a Using R fm A13 the radii are roughly fm fm and fm. b Using each of the radii in part a the areas are 163 fm2 353 fm2 and 633 fm2. c 37iR3gives 195 fm3 624 fm3 and 1499 fm3. d The density is the same since the volume and the mass are both proportional to A 1017 kg m3 see Example . e Dividing the result of part d by the mass of a nucleon the number density is fm3 x 1047m3. Æ lib - -pzB 2lib But AE hf so B 2RZ x 10-34 J s x 107 Hz _ 0 _ B T 2 x 10-27 J T a As in Example AE 2 x 10-8 eV T T x 10-7 eV. Since N and 5 are in opposite directions for a neutron the antiparallel configuration is lower energy. This result is smaller than but comparable to that found in the example for protons. .RE . c b f ---- MHz X m. h f a U j B -pZB. N and S point in the same direction for a proton. So if the spin magnetic moment of the proton is parallel to the magnetic field U 0 and if they are antiparallel U 0. So the parallel case has lower energy. The frequency of an emitted photon has a transition of the protons between the two states given by AE _ E - E- _ 2pzB f h h h _ 2 x 10-27 J T 7 ------------------------------------ X 10 HZ. x 10 34 J s c 10 8 m s . ------------ m. This is a radio wave. f x 107 Hz X b For electrons the negative charge means that the argument from part a leads to the m - -2 state antiparallel having the lowest energy since N and S point in opposite directions. So an emitted photon in a transition from one state to the other has a frequency AE E-x - En - 2 J h h h But from Eq. Pz - - - 2-00232 gn 2me 4me eB x 10-19 C T 4nme 4n x10-31 kg f x 1010 Hz so X This is a microwave. c _ 108 m s f x 1010 Hz
