Module NF2FSLT SP025

Solution
A consists of 29 protons ( Z = 29 ) & 34 neutrons
* { N = (A – Z) =( 63 – 29 ) = 34 }

Using: mass of a proton, mp is 1.007825 u
mass of a neutron, mn is 1.008665 u

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Example 194:

(a) Define the binding energy of a nucleus.
(b) Calculate the mass defect and binding energy

for

Given
(Nuclear masses , 34.968851 u;
proton1.007825 u, neutron 1.008665 u )

Solution:

(a) The binding energy of a nucleus is the energy

required to separate completely all the nucleons in

the nucleus 454

(b)

455

Example 195 ?

What is the binding energy per nucleon for
Given:

Solution

A nucleus consists of

50 protons ( Z = 50 )
& 70 neutrons ( N = (A – Z) = (120 – 50) = 70 )

456

Mass defect,
Binding Energy,

Binding Energy per nucleon,

457

Exercise 196

(1) The most abundance isotope of helium has a
nucleus whose mass is 6.6447×10–27 kg. For this
nucleus, find (a) the mass defect (b) the binding energy.

Given: mp = 1.673×10–27 kg ; mn = 1.675×10–27 kg

(2) For lead nucleus whose mass is 205.929457 u,

obtain (a) the mass defect in atomic mass unit (b) the

binding energy ( in MeV ) (c) the binding energy per

nucleon ( in MeV/nucleon ).

Answer: (1) (a) 0.0503×10–27 kg (b) 28.3 MeV 458
(2) (a) 1.741635 u (b) 1622.33 MeV (c) 7.88 MeV / nucleon

Answer
(1)

Binding Energy,

459

(2) Mass defect,

Binding Energy,
Binding Energy per nucleon,

460

Example
198

2 nuclei having the same mass number are known as
isobars. Calculate the difference in binding energy per
nucleon for the isobars and .

How do you account for the difference ?

Given:

mn = 1.008665 u
mp = 1.007825 u
mMg = 22.994127 u
mNa = 22.989770 u

461

Solution

For Mg:
Z = 12 N =( A – Z) =( 23 – 12) = 11

Mass defect,

Binding energy,

462

Binding energy per nucleon

For Na:
Z = 11 N = (A – Z) =( 23 – 11) = 12

Mass defect,

463

Binding energy,
Binding energy per nucleon
Difference in [EB/A] for both isobars :

464

Example
199
Find the energy (in MeV) released when alpha decay
converts Radium 226Ra into Radon 222Rn.

Take the masses to be:
226.025402 u for 226 Ra
222.017571 u for 222 Rn
4.002602 u for α particle

Solution
Using:

Example 200

What is the wavelength (in vacuum) of the 0.186 MeV -ray
photon emitted by radium

Solution
From:

Example 201

The half life of the radioactive nucleus Radium, is
1.6×103 year.

(a) What is the decay constant of this nucleus ?

(b) If a sample contains 3.0×1016 nuclei at t = 0 s,

determine its activity at this time.

(c) What is the activity after sample is 2.0×103 year old ?

Solution
(a) Given: T½ = 1.6 ×103 yr = 5.04576 × 1010 s

From:

(b) Given at t = 0 s, the number of radioactive nuclei
present, No = 3.0 × 1016
From:

(c) Given at t = 2 × 103 yr = 6.3072 × 1010 s, activity at that
time, A = ?

From:

Example 202
A sample of the radioisotope 131I, which has a half life of 8.04
days has an activity of 5 mCi at the time of shipment. Upon
receipt in a medical laboratory, the activity is 4.2 mCi. How
much time has elapsed between the two measurements ?

Solution Given: At t = 0 s, Ao = 5 mCi = 5×10-3 Ci
When A = 4.2 mCi = 4.2×10-3 Ci, t = ? T½ = 8.04 days

From:

From:

Example 203

A radioactive sample contains 3.5 µg of pure C, which has
a half life of 20.4 min.

(a) Determine the number of nuclei in the sample at t = 0 s.
(b) What is the activity in Becquerel of the sample initially

and after 8.0 h ?
(c) Calculate the number of radioactive nuclei remaining

after 8.0 h ?

Solution
(a) Molar mass for C = 11.0 g/mol
1 mole (11 g) of C-11 contains NA = 6.02×1023 nuclei
Thus 3.5 µg contains

Number of nuclei in the sample at t = 0s
No = 1.9155 ×1017 nuclei

(b) Given: T½ =20.4 min = 1224 s
From:

At t = 0s, the initial activity, Ao

When t = 8 h = 2.88×104 s, the activity A is
From:

(c) The number of nuclei left , N is

Ecercise 204

(a) Define the half life of a radioactive element.
(b) Carbon dating on a human remains shows the activity

of carbon-14 is 0.068 Bq per gram of carbon. The initial
activity of carbon-14 is 0.231 Bq per gram of carbon
and the half life is 5730 years. Calculate the age of the
remains.

Solution
(a) Half life of a radioactive element is the time taken for the

number of atoms in a sample of the nuclide to decreases
to half of its original value.

* Give in form of T ½ = ln 2/λ not accepted.

(b) Given:
Using:

Using:

EXAMPLE 205

What is an elementary particle?
Answer:
An elementary particle is a fundamental particle
which has no internal structure.

480

EXAMPLE 206

An electron is an example of a lepton and proton
is an example of a hadron. State a property of
hadron.

Answer:
Hadron is made up of quarks.

481

EXAMPLE 207

State the quark structure of the +, − and 0.

Answer:

K+ - usത OR up and anti −
strange quark

K− - suത OR strange and anti-up
quark

K0 - dsത OR down and anti-strange
quark

482

EXAMPLE 208

State a similarity and difference for particle and
antiparticles

Answer:
Similarity
• Both have same mass.
Difference
• They have opposite charge.

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EXAMPLE 209

State the name of the particle pത and give its properties.

Answer:

The particle is an antiproton. It has the same mass
as proton and opposite charge.

484