Nuclear equations: notes and questions PDF

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Nuclear equations: notes and questions...

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Episode 512: Nuclear equations Nuclide notation Revision of nuclide notation:

A X Z

X = Symbol for the element A = mass or nucleon number (number of protons and neutrons) Z = charge or atomic number (number of protons) N = neutron number and is related by A = Z + N Hydrogen has different symbols for its two most common isotopes after for tritium

: D for deuterium

2 D 1

and T

3 T 1

Other isotopes you may have heard of are the isotopes of carbon uranium

1 H 1

235 U 92

and

12 C 6

and

14 C 6

238 U 92

An N-Z plot (Segrè plot). Neutron number

n

p + - + 

-

A

+ B p

n + + + 

Proton number

TAP 512-1: Nuclide notation A periodic table may be needed 1 Write down the nuclear notation (a) an alpha particle

(AZ X) for:

(b) a proton (c) a hydrogen nucleus (d) a neutron (e) a beta particle (f) a positron.

1

, and the isotopes of

2 Write down the nuclear notation

(Z X) for A

(a) carbon 13 (b) nitrogen 14 (c) neon 22 (d) tin 118 (e) iron 54

TAP 512-2: Decay processes Radioactive decay processes  decay Z N

Z–2 N–2 proton number Z



2 fewer protons 2 fewer neutrons

– decay –

Z N Z+ 1 N–1

 proton number Z

1 more proton 1 less neutron

+ decay + Z–1 N+1 Z N proton number Z

 1 less proton 1 more neutron

decay



Z N

proton number Z

same protons and neutrons

2

Examples of decay equations The  source used last week was americium-241, 90 Sr 38

-

 source strontium-90,

The underlying process is:

→

1 n 0

90 39

→

Np +

1 p 1

+

0 e −1

0 e −1

+

+

241 Am 95

237 93

→

Np + 4 He 2

0 ν 0

0 ν 0

neutron → proton + electron + anti-neutrino  sources are cobalt-60. The  radiation comes from the radioactive daughter nickel-60 decay of thecobalt-60

60 Co 27

60 Ni 28

of the 

. The nickel-60 is formed in an ‘excited state’ and so almost immediately

loses the energy by emitting a   ray. They are only emitted after an  or  decay, and all such  rays have a well-defined energy. 60 Co 27

→

60 Ni 28

+

0 −1

e+

0 ν 0

+ γ

(So a cobalt-60 source which is a pure gamma emitter must be designed so that betas are not emitted. How? – (by encasing in metal which is thick enough to absorb the betas but which still allows gammas to escape.) External reference This activity is taken from Advancing Physics chapter 18, 90O

TAP 512-3: Practice with nuclear equations 1

The isotope

235

U decays into another element, emitting an alpha particle. What is the element?

This element decays, and the next, and so on until a stable element is reached. The complete 235 list of particles emitted in this chain is: 92 U→[ α , β , α , β , α , α , α , β , α , β ]→X What is the stable element X? (You could write down each element in the series, but there is a quicker way. Look at the number of α and β .)

2

The following fission reaction can take place in a nuclear reactor: 235 1 137 95 1 U + 0n → 55Cs + [ ]Rb + [ ]0 n 92 Complete the equation, showing how many neutrons are produced in the reaction. What is the significance of the number of neutrons produced?

Why are the products of the reaction, caesium-137 and rubidium-95, likely to be radioactive? What type of decay are these isotopes likely to show?

3

Boron absorbs neutrons with results as follows:

10 B 5

+ 10 n →

7 Li 3

+

4 α 2

Why is boron suitable for use in a control rod in a nuclear reactor?

4

When the isotope 27 Al is irradiated with alpha particles, the products from each aluminium 13 nucleus are a neutron, and a nuclide that emits positrons to give the stable isotope nuclear equations for these two processes.

3

30 Si 14

. Write

5

6

Complete the following nuclear equations. In each case describe the decay process: a) 131 I → Xe + −10 e + 00 ν 53 b)

67 Ga 31

c)

11 C 6

d)

99m Tc 43

+

0 −1e

7 B 3

→

→ Zn +

+ 01e +

0 ν 0

0 ν 0

→ Tc + γ

m means the isotope is in a metastable state

The Manhattan Project, the development of the atomic bomb, led to the discovery of the transuranic elements (elements beyond uranium in the periodic table). Plutonium, element 94, is formed by the bombardment of uranium-238 with neutrons. The nuclear equations are: 238 U 92

+

239 U 92

→

239 Np 93

1 n 0

239 U 92

→

239 Np 93 239 Pu 94

→

0 e −1

+ +

0 e −1

+ +

0 ν 0 0 ν 0

Complete the following nuclear equation for the formation of americium: 239 Pu 94

+ 2 10 n →

[] Am []

+

0 e −1

+

0 ν 0

Curium is produced if plutonium-239 is bombarded with alpha particles. If the curium isotope is 242 Cm , complete the equation 96 235 Pu 92

4 2

+

α →

If curium is made the target for alpha particle bombardment californium is produced. Complete the nuclear equation to find the atomic number of californium: 242 Cm 96

+

4 α 2

→

235 Cf []

+

1 n 0

By firing heavier particles such as carbon or boron ions at the target materials heavier elements can be synthesised. Complete the nuclear equation (Lw is lawrencium) 252

Cf + 95 B→ Lw +4 01 n .

252 Cf [ ]

+

9 B 5

→

[ ] Lw [ ]

+ 4 10 n

One of the transuranic elements is commonly found in the home. Which is this and where is it used?

Answers Answers - TAP 512-1: Nuclide notation 1

(a) an alpha particle (b) a proton

1 1

H or

4 He 2

1 1

(c) a hydrogen nucleus (d) a neutron

p 1 1

H

1 n 0

(e) a beta particle

0 e −1

4

2

(f) a positron.

0 e +1

(a) carbon-13

13 C 6

(b) nitrogen-14

14 N 7

22 10 Ne

(c) neon-22 (d) tin-118

118 Sn 50

(e) iron-54

54 26 Fe

Answers and worked solutions TAP 512-3: Practice with nuclear equations 1.

) and four beta The complete decay chain involves the loss of seven alpha particles ( 4 2α particles (

0 e −1

). This represents a loss of 7  4, i.e. 28, in mass number and (7  2 – 4), i.e.

10, in atomic number. X is therefore an isotope with mass number (235 – 28), i.e. 207, and atomic number (92 – 10), i.e. 82. This is lead, 2.

235 U 92

208 Pb 82

.

+ 01 n→ 137 Cs + 95 Rb + 3 10 n . 55 37

The reaction produces more neutrons than it absorbs, this will cause a ‘chain reaction’. To see why the products of the reaction are likely to be radioactive you need to consult the plot of neutron number against atomic number for the known stable nuclei. For elements with atomic numbers up to about 30 the number of neutrons in the nucleus is the same as the number of protons if the nucleus is stable. For higher atomic numbers the ratio of neutrons to protons gradually increases to 1.5. Look at the position of both 137 and 95 on the 55 Cs 37 Rb plot and you will see that they both have a considerable excess of neutrons. They are therefore likely to be radioactive. To become more stable the nuclides need to decrease the neutron to proton ratio. The emission of a beta particle does this, increasing the number of protons by one and decreasing the number of neutrons by one. These isotopes are therefore likely to decay by emitting beta particles. 3.

4.

5.

6.

When boron captures a neutron it is transformed into a stable isotope. If the control rods are pushed into the reactor more neutrons are absorbed, causing the chain reaction to slow down. If they are pulled out the chain reaction will proceed more vigorously. 27 30 Al + 24 α = 15 P + 10 n 13 30 30 P = 14 Si + 10 e 15 131 I → 131 Xe + −10 e beta decay 54 53 67 Ga + −10 e → 67 Zn electron capture 30 31 11 11 0 6 C → 5 B + 1 e positron emission 99 m 99 43 Tc→ 43 Tc nuclear rearrangement 239 Pu + 2 10 n→ 241 Am + −10 e 95 94 239 Pu + 24 α→ 242 Cm + 01 n 96 94 242 Cm + 24 α →245 Cf + 01 n 98 96 252 257 Cf + 95 B→ 103 Lw + 4 01 n . 98

Americium is commonly found in homes where it is used in smoke detectors. External reference This activity is taken from Advancing Physics chapter 18, 210S

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