Page 222 - Physics Form 5 KSSM_Neat

Solving problems involving Nuclear Energy due to Radioactive Decay and
Nuclear Reactions
Example 1 LET’S ANSWER
LET’S ANSWER
The equation below shows radium-226 decaying into radon-222
by emitting alpha particle.
KEMENTERIAN PENDIDIKAN MALAYSIA
4
226 Ra 222 Rn + He

88 86 2
222
226

Given that the mass of Ra is 226.54 amu, Rn is 222.018 amu and http://bit.

86
88
4 He is 4.003 amu, calculate the nuclear energy released. ly/2Q9FTQs
2
[1 amu = 1.66 × 10 –27 kg and speed of light in vacuum, c = 3.00 × 10 m s ]
8
–1
Solution
Step 1: Step 2: Step 3: Step 4:
Identify the Identify the Identify the formula Solve the problem
problem information given that can be used numerically
1 Nuclear energy that is released 3 Nuclear energy released, E = mc 2

2 Mass 226 Ra = 226.54 amu 4 Mass defect, m = 226.54 – (222.018 + 4.003)
88 = 0.519 amu
Mass 222 Rn = 222.018 amu Mass defect, m (in kg) = 0.519 × 1.66 × 10
–27
86
4
–28
Mass He = 4.003 amu = 8.6154 × 10 kg

2 E = mc 2
Speed of light in vacuum, –28 8 2
c = 3.00 × 10 m s –1 = 8.6154 × 10 × (3.00 × 10 )
8
–11
1 amu = 1.66 × 10 kg = 7.75 × 10 J
–27
Example 2
The Sun is the source of energy to the Earth. This energy is produced by the nuclear fusion in
the core of the Sun as shown in the following equation:
3
1
4
2 H + H He + n + energy

1 1 2 0
Calculate the nuclear energy released in joules.
2
4
3
[Mass of H = 2.014 amu, mass of H = 3.016 amu, mass of He = 4.003 amu,
1
1
2
1
mass of n = 1.009 amu, 1 amu = 1.66 × 10 kg and
–27
0
the speed of light in a vacuum, c = 3.0 × 10 m s ]
8
–1
Solution
–27
Mass defect, m Thus, m = 0.018 × 1.66 × 10 kg
–29
= (2.014 + 3.016) – (4.003 + 1.009) m = 2.988 × 10 kg
= 5.030– 5.012 E = mc 2
8 2
= 0.018 amu = 2.988 × 10 × (3.0 × 10 )
–29
= 2.69 × 10 J
–12
212 LS 6.2.3