Author : Khairi

NUCLEAR ENERGY

FORMULA

E = mc2

E = Energy release,J or eV

m = mass defect,kg (*Always in positive number)

c = speed of light = 3 x 108 ms-1

Author : Khairi

E = mc2

m = mass defect

Mass defect = the differences of mass before and after reaction

Atomic mass is too small,therefore it is measured inAtomic mass unit (a.m.u)

1 a.m.u = 1/12 of mass of carbon-12 atom.

= 1/12 x 1.993 x 10-27 kg

= 1.66 x 10-27 kg

NUCLEAR ENERGY

Author : Khairi

EXERCISE

Calculate the mass of atoms below in kg

1. Radium –225 = 222.0254 a.m.u2. Radon – 222 = 222.0175 a.m.u3. Helium-4 = 4.0026 a.m.u

1 a.m.u = 1.66 x 10-27 kg

1. 222.0254 x 1.66 x 10-27 = 3.68562164 x 10-25 kg

2. 222.0175 x 1.66 x 10-27 = 3.6854905 x 10-25 kg

3. 4.0026 x 1.66 x 10-27 = 6.44316 x 10-27 kg

NUCLEAR ENERGY

Author : Khairi

Nuclear equation of Radium-226 decay as follows;

CALCULATING NUCLEAR ENERGY

Ra22688 Rn222

86+ He4

2+ energy

Calculate the total nuclear energy released in Joule and eV.

Given : Ra-226 = 226.0254 a.m.u Rn-222 = 222.0175 a.m.u He-4 = 4.0026 a.m.u

E = mc2

Mass before = 226.0254 a.m.u

Mass after = 222.0175 + 4.0026

= 226.0201 a.m.u

Mass defect = 226.0254 - 226.0201

= 0.0053 a.m.u

= 0.0053 x 1.66 x 10-27

= 8.798 x 10-30 kg

= 8.798 x 10-30 x (3 x 108)2

= 7.92 x 10-13 J

NUCLEAR ENERGY

Author : Khairi

Nuclear equation of Radium-226 decay as follows;

CALCULATING NUCLEAR ENERGY

Ra22688 Rn222

86+ He4

2+ energy

Calculate the total nuclear energy released in Joule and eV.

Given : Ra-226 = 226.0254 a.m.u Rn-222 = 222.0175 a.m.u He-4 = 4.0026 a.m.u

E = mc2

= 8.798 x 10-30 x (3 x 108)2

= 7.92 x 10-13 J

NUCLEAR ENERGY

E = E/e

E = 7.92 x 10-13/1.60 x 10-19

= 4.88 x 106 eV

Author : Khairi

Below is a nuclear equation for nuclear reaction;

EXERCISE

U23592 Xe140

54 n10n1

0 Sr9438

+ + energy+ X +

Given : U-235 = 235.043924 a.m.u n = 1.008665 a.m.u Xe-140 = 139.921620 a.m.u Sr-94 = 93.915367 a.m.u

1. Determine the value of X

2. Calculate the total nuclear energy released in Joule and eV

NUCLEAR ENERGY

Author : Khairi

Nucleon number

235 + 1 = 140 + 94 + 1X

236 = 234 + 1XX = 2

EXERCISE

NUCLEAR ENERGY

Below is a nuclear equation for nuclear reaction;

1. Determine the value of X

U23592 Xe140

54 n10n1

0 Sr9438

+ + energy+ X +

Given : U-235 = 235.043924 a.m.u n = 1.008665 a.m.u Xe-140 = 139.921620 a.m.u Sr-94 = 93.915367 a.m.u

Author : Khairi

2. Calculate the total nuclear energy released in Joule and eV.

Mass before = 235.043924 + 1.008665= 236.052589 a.m.u

Mass after = 139.921620 + 93.915367 + 2(1.008665)= 235.854317 a.m.u

NUCLEAR ENERGYEXERCISE

Below is a nuclear equation for nuclear reaction;

U23592 Xe140

54 n10n1

0 Sr9438

+ + energy+ 2 +

Given : U-235 = 235.043924 a.m.u n = 1.008665 a.m.u Xe-140 = 139.921620 a.m.u Sr-94 = 93.915367 a.m.u

Author : Khairi

Mass defect = 236.052589 – 235.854317

= 0.198272 a.m.u= 0.198272 x 1.66 x 10-27

= 0.32913152 x 10-27 kg

E = mc2

= 0.32913152 x 10-27 x ( 3 x 108)2

= 2.962 x 10-11 J

NUCLEAR ENERGYEXERCISE

Below is a nuclear equation for nuclear reaction;

= 1.9 x 108 eV

U23592 Xe140

54 n10n1

0 Sr9438

+ + energy+ 2 +

Author : Khairi

a. Complete the equation for the reaction above

4

86

b. The nuclear reaction of on nucleus of Radium-226 experience a mass defect of 8.61818 x 10-30 kg.Calculate the energy released in the nuclear reaction.

E = mc2

= 8.61818 x 10-30 x (3 x 108)2

= 7.76 x 10-15 J

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