Nulear Chemistry Mizzou University of Missouri核化學(xué)美國(guó)密蘇里堪薩斯大學(xué)

1、1nuclear chemistrychapter 19copyright the mcgraw-hill companies, inc. permission required for reproduction or display.2xazmass number atomic numberelement symbolatomic number (z) = number of protons in nucleus mass number (a) = number of protons + number of neutrons = atomic number (z) + number of n

2、eutronsaz1p11h1orproton1n0neutron0e-10b-1orelectron0e+10b+1orpositron4he24a2ora particle11100-10+142review3balancing nuclear equations1. conserve mass number (a). the sum of protons plus neutrons in the products must equal the sum of protons plus neutrons in the reactants.1n0u23592+cs13855rb96371n0+

3、 2235 + 1 = 138 + 96 + 2x12. conserve atomic number (z) or nuclear charge. the sum of nuclear charges in the products must equal the sum of nuclear charges in the reactants.1n0u23592+cs13855rb96371n0+ 292 + 0 = 55 + 37 + 2x04example519.1balance the following nuclear equations (that is, identify the

4、product x):(a) (b) 2122088482po pb + x1371375556cs ba + xexample619.1solution(a) the mass number and atomic number are 212 and 84, respectively, on the left-hand side and 208 and 82, respectively, on the right-hand side. thus, x must have a mass number of 4 and an atomic number of 2, which means tha

5、t it is an particle. the balanced equation isstrategy in balancing nuclear equations, note that the sum of atomic numbers and that of mass numbers must match on both sides of the equation.2122088482po pb + 42aexample7(b) in this case, the mass number is the same on both sides of the equation, but th

6、e atomic number of the product is 1 more than that of the reactant. thus, x must have a mass number of 0 and an atomic number of -1, which means that it is a particle. the only way this change can come about is to have a neutron in the cs nucleus transformed into a proton and an electron; that is, (

7、note that this process does not alter the mass number). thus, the balanced equation is19.111001-1n p + 1371375556-1cs ba + 0example8check note that the equation in (a) and (b) are balanced for nuclear particles but not for electrical charges. to balance the charges, we would need to add two electron

8、s on the right-hand side of (a) and express barium as a cation (ba+) in (b).19.19nuclear stabilitycertain numbers of neutrons and protons are extra stablen or p = 2, 8, 20, 50, 82 and 126like extra stable numbers of electrons in noble gases (e- = 2, 10, 18, 36, 54 and 86)nuclei with even numbers of

9、both protons and neutrons are more stable than those with odd numbers of neutrons and protonsall isotopes of the elements with atomic numbers higher than 83 are radioactiveall isotopes of tc and pm are radioactive1011n/p too largebeta decayxn/p too smallpositron decay or electron capturey12nuclear s

10、tability and radioactive decaybeta decay14c 14n + 0b67-140k 40ca + 0b1920-11n 1p + 0b01-1decrease # of neutrons by 1increase # of protons by 1positron decay11c 11b + 0b 65+138k 38ar + 0b 1918+11p 1n + 0b10+1increase # of neutrons by 1decrease # of protons by 113electron capture decayincrease number

11、of neutrons by 1decrease number of protons by 1nuclear stability and radioactive decay37ar + 0e 37cl 1817-155fe + 0e 55mn 2625-11p + 0e 1n10-1alpha decaydecrease number of neutrons by 2decrease number of protons by 2212po 4he + 208pb84282spontaneous fission252cf 2125in + 21n9849014nuclear binding en

12、ergy is the energy required to break up a nucleus into its component protons and neutrons.nuclear binding energy + 19f 91p + 101n9109 x (p mass) + 10 x (n mass) = 19.15708 amude = (dm)c2dm= 18.9984 amu 19.15708 amudm = -0.1587 amude = -2.37 x 10-11jde = -0.1587 amu x (3.00 x 108 m/s)2= -1.43 x 1016

13、amu m2/s2using conversion factors: 1 kg = 6.022 x 1026 amu1 j = kg m2/s215= 2.37 x 10-11 j19 nucleons= 1.25 x 10-12 j/nucleonbinding energy per nucleon = binding energynumber of nucleonsde = (-2.37 x 10-11j) x (6.022 x 1023/mol)de = -1.43 x 1013j/molde = -1.43 x 1010kj/molnuclear binding energy = 1.

14、43 x 1010kj/mol 16nuclear binding energy per nucleon vs mass numbernuclear stabilitynuclear binding energynucleonexample1719.2the atomic mass of is 126.9004 amu. calculate the nuclear binding energy of this nucleus and the corresponding nuclear binding energy per nucleon.12753iexample1819.2strategy

15、to calculate the nuclear binding energy, we first determine thedifference between the mass of the nucleus and the mass of all the protons and neutrons, which gives us the mass defect. next, we apply equation (19.2) e = (m)c2.solutionthere are 53 protons and 74 neutrons in the iodine nucleus. the mas

16、s of 53 atom is 53 x 1.007825 amu = 53.41473 amuand the mass of 74 neutrons is74 x 1.008665 amu = 74.64121 amu1h1example1919.2therefore, the predicted mass for is 53.41473 + 74.64121 = 128.05594 amu, and the mass defect is m = 126.9004 amu - 128.05594 amu= -1.1555 amuthe energy released ise = (m)c2=

17、 (-1.1555 amu) (3.00 x 108 m/s)2= -1.04 x 1017 amu m2/s212753iexample2019.2lets convert to a more familiar energy unit of joules. recall that 1 j = 1 kg m2/s2. therefore, we need to convert amu to kg:thus, the nuclear binding energy is 1.73 x 10-10 j . the nuclear binding energy per nucleon is obtai

18、ned as follows:2172232-10-102amu m1.00 g1 kg-1.040鬃1000 gs6.02210 amukg m = -1.7310 = -1.7310 jsde-10 1.7310 j= = 127 nucleons-121.3610 j / nucleon21kinetics of radioactive decayn daughterrate = lnn = the number of atoms at time tn0 = the number of atoms at time t = 0l is the decay constantntlnn0-lt

19、=tl0.6932223radiocarbon dating14n + 1n 14c + 1h716014c 14n + 0b + n67-1t = 5730 yearsuranium-238 dating238u 206pb + 8 4a + 6 0b92-1822t = 4.51 x 109 years24nuclear transmutation14n + 4a 17o + 1p728127al + 4a 30p + 1n13215014n + 1p 11c + 4a7162example25write the balanced equation for the nuclear reac

20、tion where d represents the deuterium nucleus (that is, ).19.3h2156542625fe(d,)mnexample26strategy to write the balanced nuclear equation, remember that the first isotope is the reactant and the second isotopeis the product. the first symbol in parentheses (d) is the bombarding particle and the seco

21、nd symbol in parentheses () is the particle emitted as a result of nuclear transmutation.19.35626fe5425mnexample2719.3solution the abbreviation tells us that when iron-56 is bombarded with a deuterium nucleus, it produces the manganese-54 nucleus plus an particle. thus, the equation for this reactio

22、n ischeckmake sure that the sum of mass numbers and the sum of atomic numbers are the same on both sides of the equation.562454261225fe + h + mn28nuclear transmutation29nuclear fission235u + 1n 90sr + 143xe + 31n + energy92543800energy = mass 235u + mass n (mass 90sr + mass 143xe + 3 x mass n ) x c2

23、energy = 3.3 x 10-11j per 235u= 2.0 x 1013 j per mole 235ucombustion of 1 ton of coal = 5 x 107 j30nuclear fission235u + 1n 90sr + 143xe + 31n + energy92543800representative fission reaction3132nuclear fissionnuclear chain reaction is a self-sustaining sequence of nuclear fission reactions.the minimum mass of fissionable material required to generate a self-sustaining nuclear chain reaction is the critical mass.33schematic of an atomic bomb34schematic diagram of a nuclear reactoru3o8refueling35chemistry in action: natures own fission reac