*

is defined as the spontaneous disintegration of certain atomic nuclei accompanied by the emission of alpha particles, beta particles or gamma radiation.

Radioactivity

Radioactive decay

Radioactivity is a phenomenon in which an unstable nuclei undergoes spontaneous decay as a result of which a new nucleus is formed and energy in the form of radiation is released

The radioactive decay is a spontaneous reaction that is unplanned, cannot be predicted and independent of physical conditions (such as pressure, temperature) and chemical changes.This reaction is random reaction because the probability of a nucleus decaying at a given instant is the same for all the nuclei in the sample.Radioactive radiations are emitted when an unstable nucleus decays. The radiations are alpha particles, beta particles and gamma-rays.

*

SF027

Alpha particle ()

An alpha particle consists of two protons and two neutrons.It is identical to a helium nucleus and its symbol is It is positively charged particle and its value is +2e with mass of 4.002603 u.When a nucleus undergoes alpha decay it loses four nucleons, two of which are protons, thus the reaction can be represented by general equation below:

*

OR

(Parent)

( particle)

(Daughter)

Alpha particles can penetrate a sheet of paper.

SF027

*

Examples of decay :

parent

daughter

particle

*

Two types :

a) Beta minus , -

b) Beta plus , +

A beta particle has the same mass and charge as an electron.

Beta particles can penetrate a few mm of Al and their velocity is high (v ~ c).

Beta particle ()

*

*

Beta minus ( )-negatively charge.

Also called as negatron or electron.

Symbol;

- or

It is produced when one of the neutrons in the parent nucleus decays into a proton, an electron and an antineutrino.

massless, neutral

*

*

In beta-minus decay, an electron is emitted, thus the mass number does not charge but the charge of the parent nucleus increases by one as shown below :

(Parent)

( particle)

(Daughter)

Examples of minus decay :

*

*

Beta plus (+)- positively charge.

Also called as positron or antielectron.

Symbol;

+ or

It is produced when one of the protons in the parent nucleus decays into a neutron, a positron and

a neutrino.

massless,neutral

*

*

In beta-plus decay, a positron is emitted, this time the charge of the parent nucleus decreases by one as shown below :

(Parent)

(Positron)

(Daughter)

Example of plus decay :

*

Gamma ray ()

Gamma rays are high energy photons (electromagnetic radiation).Emission of gamma ray does not change the parent nucleus into a different nuclide, since neither the charge nor the nucleon number is changed.A gamma ray photon is emitted when a nucleus in an excited state makes a transition to a ground state.Examples of decay are :It is uncharged (neutral) ray and zero mass.The differ between gamma-rays and x-rays of the same wavelength only in the manner in which they are produced; gamma-rays are a result of nuclear processes, whereas x-rays originate outside the nucleus.

*

Gamma ray

SF027

Comparison of the properties between alpha particle, beta particle and gamma ray.

Table 1 shows the comparison between the radioactive radiations.

*

+2e

1e OR +1e

0 (uncharged)

Yes

Yes

No

Strong

Moderate

Weak

Weak

Moderate

Strong

Yes

Yes

Yes

Yes

Yes

Yes

Table 1

AlphaBetaGammaChargeDeflection by electric and magnetic fieldsIonization powerPenetration powerAbility to affect a photographic plateAbility to produce fluorescence

SF027

Decay constant ()

Law of radioactive decay states:

For a radioactive source, the decay rate is directly

proportional to the number of radioactive nuclei N remaining in the source.

i.e.

Rearranging the eq. (1):

Hence the decay constant is defined as the probability that a radioactive nucleus will decay in one second. Its unit is s1.

*

Negative sign means the number of remaining nuclei decreases with time

Decay constant

(1)

SF027

The decay constant is a characteristic of the radioactive nuclei.Rearrange the eq. (1), we get

At time t=0, N=N0 (initial number of radioactive nuclei in the sample) and after a time t, the number of remaining nuclei is N. Integration of the eq. (2) from t=0 to time t :

*

Exponential law of radioactive decay

(2)

(3)

SF027

*

The number of nuclei N as function of time t

Half-life is the time required for the number of radioactive nuclei to decrease to half the original number (No)

*

From

Hence

*

The units of the half-life are second (s), minute (min), hour (hr), day (d) and year (y). Its unit depend on the unit of decay constant.Table 2 shows the value of half-life for several isotopes.

*

Table 2

IsotopeHalf-life4.5 109 years1.6 103 years138 days 24 days3.8 days20 minutes

SF027

Activity of radioactive sample (A)

is defined as the decay rate of a radioactive sample.Its unit is number of decays per second.Other units for activity are curie (Ci) and becquerel (Bq) S.I. unit.Unit conversion:Relation between activity (A) of radioactive sample and time t :From the law of radioactive decay :

and definition of activity :

*

SF027

Thus

*

and

Activity at time t

Activity at time, t =0

and

(4)

SF027

*

A radioactive nuclide A disintegrates into a stable nuclide B. The half-life of A is 5.0 days. If the initial number of nuclide A is 1.01020, calculate the number of nuclide B after 20 days.

Solution :

The decay constant is given by

The number of remaining nuclide A is

The number of nuclide A that have decayed is

Therefore the number of nuclide B formed is

Example 1 :

SF027

*

80% of a radioactive substance decays in 4.0 days. Determine

i. the decay constant,

ii. the half-life of the substance.

Example 2 :

SF027

*

Solution :

At time

The number of remaining nuclei is

i. By applying the exponential law of radioactive decay, thus the

decay constant is

ii. The half-life of the substance is

SF027

*

A thorium-228 isotope which has a half-life of 1.913 years decays by emitting alpha particle into radium-224 nucleus. Calculate

a. the decay constant.

b. the mass of thorium-228 required to decay with activity of

12.0 Ci.

c. the number of alpha particles per second for the decay of 15.0 g

thorium-228.

(Given the Avogadro constant, NA =6.02 1023 mol1)

Solution :

a. The decay constant is given by

Example 3 :

SF027

*

Solution :

b. By using the unit conversion ( Ci decay/second ),

the activity is

Since then

If 6.02 1023 nuclei of Th-228 has a mass of 228 g thus

3.86 1019 nuclei of Th-228 has a mass of

SF027

*

Solution :

c. If 228 g of Th-228 contains of 6.02 1023 nuclei thus

15.0 g of Th-228 contains of

Therefore the number of emitted alpha particles per second is

given by

Ignored it.

SF027

*

A sample of radioactive material has an activity of 9.00 x

1012 Bq. The material has a half-life of 80.0 s. How long

will it take for the activity to fall to 2.00 x 1012 Bq ?

Solution

Example 4 :

*

N = 25% , No = 100%

t =34.6 min

(1.72 h)

Example 5 :

*

Good luck

For

2nd semester examination

SF027

He

4

2

4

2

X

A

Z

+

Y

4

2

-

-

A

Z

Q

He

4

2

Q

+

+

He

Pb

Po

4

2

214

82

218

84

Q

+

+

He

Ra

Th

4

2

226

88

230

90

Q

+

+

He

Rn

Ra

4

2

222

86

226

88

Q

+

+

He

Th

U

4

2

234

90

238

92

e

0

1

0

1

-

-

or

b

Q

e

Pa

Th

0

1

234

91

234

90

+

+

-

Q

e

U

Pa

0

1

234

92

234

91

+

+

-

Q

e

Po

Bi

0

1

214

84

214

83

+

+

-

X

A

Z

Y

A

1

Z

+

e

0

1

-

e

0

1

0

1

or

b

Q

v

e

C

N

+

+

+

0

1

12

6

12

7

Y

A

1

Z

-

e

0

1

+

+

*

He

Pb

Po

4

2

214

82

218

84

+

+

-

*

e

U

Pa

0

1

234

92

234

91

+

*

Ti

Ti

208

81

208

81

dt

dN

N

dt

dN

N

dt

dN

nuclei eradioactiv remaining ofnumber

ratedecay

N

dt

dN

dt

N

dN

tN

N

dt

N

dN

0

0

tN

N

tN

0

0

ln

t

N

N

0

ln

t

eNN

0

t

eNN

0

t

0

eNN

2

1

T

0

0

eN

2

N

2

1

T

e2

2

1

T

e

2

1

2

1

T

e2

lnln

6930

2

T

2

1

.ln

2

2

1

o

N

N

T

t

=

=

and

When

U

238

92

Po

210

884

Ra

226

88

Bi

214

83

Rn

222

86

Th

234

90

dt

dN

secondper decay 1Bq 1

N

dt

dN

dt

dN

A

secondper decays1073Ci1

10

.

00

NA

NA

t

eNN

0

t

eNA

0

t

eN

0

t

eAA

0

Q

B

A

+

days

20

;

10

1.0

days;

0

.

5

20

0

2

/

1

=

=

=

t

N

T

t

e

N

N

l

-

=

0

t

e

N

N

l

-

=

0

l

2

ln

2

/

1

=

T

(

)

60

60

24

365

1.913

y

913

.

1

2

/

1

=

=

T

s

10

03

.

6

7

=

NA

A

N

secondper decays1073Ci1

10

.

N

dt

dN

A

s 174

o

o

A

A

tt

A

A

ln

ln

T

2ln

2

1

2/1

2ln

t

t

o

t

e

A

A

eAA

0

t

eNN

0

2/1

2ln

t