z a nucleons, i.e. ( a , which means that it has a total...

FLAP P9.1 Introducing atomic nucleiCOPYRIGHT © 1998 THE OPEN UNIVERSITY S570 V1.1

F1

The nucleus ZAX is the nucleus of an atom of chemical element X. It has atomic number Z, which means that it

has Z protons. It has mass number A, which means that it has a total of A nucleons, i.e. (A − Z) neutrons.So the nucleus 10

22Ne has 10 protons and 12 neutrons and the nucleus 4094Zr has 40 protons and 54 neutrons.


Properties of some common nuclides

Z Element Symbol A Atomic mass/u Relativeabundance/%

1 hydrogen H 1 1.0071825 99.985

2 2.0141102 0.015

2 helium He 4 4.0021603 99.999

79 gold Au 197 196.9661560 100

82 lead Pb 204 203.9731037 1.4

206 205.9741455 24.1

207 206.9751885 22.1

208 207.9761641 52.4

83 bismuth Bi 209 208.9801388 100

* The neutron, in the free state, is not stable; it has a mean lifetime of approximately151min. It has a mass of 1.00816651u. In contrast, a free proton is stable and has a massof 1.00712761u.

F2

The nucleus 82208Pb has 82

protons and 126 neutrons;from the table its mass is207.97616411u. The total massof the free neutrons andprotons is

82 × 1.00718251u + 126 ×

1.00816651u = 209.7331441u

(again using Table 1). Themass defect is, therefore,209.7331441u − 207.97616411u= 1.75681u.


The binding energy in MeV (from E = mc2) is therefore given by:

binding energy = 1.7568 × 931.5021(MeV/c2) × c2 = 1636.51MeV


Study comment Having seen the Fast track questions you may feel that it would be wiser to follow the normal routethrough the module and to proceed directly to Ready to study? in Subsection 1.3.

Alternatively, you may still be sufficiently comfortable with the material covered by the module to proceed directly to theClosing items.

If you have completed both the Fast track questions and the Exit test, then you have finished the module and may leave ithere.


R1

Like charges (i.e. both positive or both negative) repel each other.


R2

The energy Eel transferred when a charge q is accelerated through a potential difference V is E el = qV.An electron has a charge of magnitude 1.602 × 10−19

1C so the energy transferred to the electron in accelerating itthrough 51V is Eel = 51V × 1.602 × 10−19

1C = 8.010 × 10−191J. This is equal to the kinetic energy acquired by the

electron.


R3

Newton’s law of universal gravitation states that every body attracts every other body with a force proportionalto the mass of each and inversely proportional to the square of the distance of separation. For bodies with massesm1 and m2 separated by a distance r the magnitude of the force Fgrav is:

Fgrav = Gm1m2

r2

where G is the universal gravitational constant.

(If you feel unsure about any of the terms used in Questions R1 to R3, consult the Glossary.)


T1

1428Si . Strictly speaking the notation Z

AX is not as brief as it could be because the number Z can refer only to one

chemical symbol X. For instance 1428Si can refer only to silicon since Z = 14. Therefore either Z or X could be

left out of the notation. But redundant chemical information is not necessarily a bad thing; here it reminds thereader of the chemical element with this value of Z and checks that the information is consistent. This notationhas become standard.


T2

The nucleus of the isotope 3068Zn has (a) 30 protons, and (b) 68 − 30 = 38 neutrons. (c) The charge on the nucleus

is 30 × (+1.6 × 10−191C) = +4.8 × 10−18

1C. (d) The neutral atom has the same number of electrons as protons.Therefore it contains 30 electrons. (e) Nucleus X has 31 protons, whereas zinc always has 30, therefore X cannotbe the chemical element zinc1—1it is actually the element gallium.


T3

The atom of 612C has six electrons of total mass 6 × (9.110 × 10−31

1kg) = 5.4654 × 10−301kg. This mass in atomic

mass units u is:

5. 4654 × 10−30 kg1. 6606 × 10−27 kg u−1

= 3. 291 × 10−3 u

The mass of the 612C nucleus is the mass of the atom minus the mass of the electrons, i.e.

(12 − 3.291 × 10−3)1u = 11.99617091u

The percentage is therefore (3.291 × 10−3/12) × 100 = 0.027%.

Comment This is for a light nucleus, with an equal number of protons and neutrons. For heavier nuclei, there aremore neutrons than protons in the nucleus, so the electrons contribute a smaller percentage of the atomic mass. For precisenuclear calculations, such as you will encounter later in this module, the actual masses of nuclei should be used, but for manypurposes the mass of the nucleus can be taken to be the same as the mass of the atom.


T4

From Figure 3, the point corresponding to Z = 69 is N = 100 (there is only one isotope). The neutron/proton ratiotherefore is 100/69 = 1.45. For Z = 79, N = 118 the ratio is 1.49. (The ratio increases with increasing Z, as is evident from the general trend of the points plotted.)


T5

The kinetic energy acquired by the proton is

1.602 × 10−191C × 103

1V = 1.602 × 10−161J = 103

1eV = 10−31MeV.


T6

The energy equivalent for the mass of an electron is

mec2 = 9.110 × 10−311kg × (2.998 × 108

1m1s−1)2 = 8.188 × 10−141J

Since 11eV = 1.602 × 10−191J, the energy equivalent to the mass of the electron is therefore

8.188 × 10−14 J1. 602 × 10−19 J eV−1

= 5.111 × 105 eV = 0. 5111MeV

The electron therefore has a mass of 0.51111MeV/c2 (i.e. the mass which, when multiplied by c2 gives0.51111MeV).


T7

11u = 1.6606 × 10−271kg. Using the same method as in Answer T6, we find:

1 u =1. 6606 × 10−27 kg × 2. 998 × 108 m s−1( )2

1. 602 × 10−19 J eV−1 = 9.317 × 108

1eV/c2 = 931.71MeV/c2

(A more precise figure obtained using a more precise value of c and e is 11u = 931.5021MeV/c2.)


T8

From the data in Subsection 3.2 and the definition of the above mass unit the mass defect is:

(6 × 1.0071825)1u + (6 × 1.0081665)1u − 121u = 0.0981941u

Using Answer T7 (i.e. 11u = 931.71MeV/c2), the mass defect is 92.181MeV/c2, and hence the binding energy is92.181MeV.


T9The decay may be written:

90230Th → Z

ARa + 24He

Equating the number of protons on each side gives 90 = Z + 2 and equating the total number of nucleons on eachside gives 230 = A + 4. Thus the reaction is:

90230Th → 88

226Ra + 24He

(Z = 88 confirms that the decay-product nucleus is indeed radium.)


Table 23Properties of some unstable nuclides.

Z Element Symbol A Atomic mass/ u Relativeabundance/%

Half-life

1 hydrogen H 1 1.0071825 99.985

2 2.0141102 0.015

2 helium He 4 4.0021603 99.999

88 radium Ra 222 222.0151353 381s

226 226.0251406 1.60 × 1031yr

90 thorium Th 230 230.0331131 8.0 × 1041yr

233 233.0411579 22.31min

92 uranium U 235 235.0431925 0.72 7.04 × 1081yr

238 238.0501786 99.28 4.47 × 1091yr

T10

From Answer T9, theequation is

90230Th → 88

226Ra + 24He

From Equation 4,

Q = (mass of nucleus − sum

of masses of decay

fragments) × c2 (Eqn 4)

the additional data in Table 2gives the mass-equivalent ofthe Q-value of:

230.03311311u − (226.0251406 + 4.0021603)1u = 0.00511221u


This means that Q = 0.00511221u × 931.5021MeV/u = 4.7711MeV. The decay equation is therefore

90230Th → 88

226Ra + 24He + 4. 771MeV


Properties of some unstable nuclides.

Z Element Symbol A Atomic mass/ u Relativeabundance/%

Half-life

1 hydrogen H 3 3.0161050 12.31yr

2 helium He 4 4.0021603 99.999

4 beryllium Be 8 8.0053 ~3 × 10−161s

6 carbon C 14 14.0031242 57301yr

19 potassium K 40 39.9631999 0.01 1.28 × 1091yr

53 iodine I 131 130.9061119 8.041days

76 osmium Os 186 185.9531852 1.6 2 × 10151yr

78 platinum Pt 190 189.9591937 0.013 6 × 10111yr

84 polonium Po 218 218.0081930 3.051min

86 radon Rn 218 218.0051580 0.0351s

222 222.0171574 3.821days

87 francium Fr 221 221.0141183 4.81min

T11

78190Pt → 76

186Os + 24He

The mass-equivalent of theQ-value is equal to:

189.95919371u −

(185.9531852 + 4.0021603)1u

= 0.00314821u

This is positive, so thedecay is energeticallypossible.

z a nucleons, i.e. ( a , which means that it has a total...

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