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Physics: Nuclear Physics - Radioactivity, Decay Equations, and Half-Life

Grade 8IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Radioactivity is the spontaneous emission of radiation from the nucleus of an unstable atom. The aim of this process is for the nucleus to reach a more stable state.

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The atom is represented by the notation ZAX^{A}_{Z}X, where AA is the Mass Number (protons + neutrons) and ZZ is the Atomic Number (protons).

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Alpha Decay (α\alpha): The nucleus emits an alpha particle, which consists of 2 protons and 2 neutrons (a Helium nucleus, 24He^{4}_{2}He). This reduces the mass number by 4 and the atomic number by 2.

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Beta Decay (β\beta): A neutron in the nucleus turns into a proton and emits a fast-moving electron (−10e^{0}_{-1}e). The mass number remains the same, but the atomic number increases by 1.

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Gamma Decay (γ\gamma): The nucleus releases excess energy as high-frequency electromagnetic radiation. There is no change in the mass or atomic number of the element.

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Half-Life (t1/2t_{1/2}): The time taken for the number of radioactive nuclei in a sample to decrease by half, or for the activity (measured in Becquerels, BqBq) to drop to half its initial value.

📐Formulae

Alpha Decay: ZAX→Z−2A−4Y+24α\text{Alpha Decay: } ^{A}_{Z}X \rightarrow ^{A-4}_{Z-2}Y + ^{4}_{2}\alpha

Beta Decay: ZAX→Z+1AY+−10β\text{Beta Decay: } ^{A}_{Z}X \rightarrow ^{A}_{Z+1}Y + ^{0}_{-1}\beta

Number of half-lives (n): n=Total Time (T)Half-life (t1/2)\text{Number of half-lives (n): } n = \frac{\text{Total Time (T)}}{\text{Half-life } (t_{1/2})}

Remaining Amount (N): N=N0×(12)n\text{Remaining Amount (N): } N = N_{0} \times \left(\frac{1}{2}\right)^{n}

💡Examples

Problem 1:

A sample of Radium-226 has an initial mass of 160 g160\text{ g}. If its half-life is 16001600 years, calculate the mass of Radium-226 remaining after 48004800 years.

Solution:

n=48001600=3n = \frac{4800}{1600} = 3 N=160×(12)3N = 160 \times \left(\frac{1}{2}\right)^{3} N=160×18=20 gN = 160 \times \frac{1}{8} = 20\text{ g}

Explanation:

First, calculate the number of half-lives (nn) that have passed by dividing the total time by the half-life duration. Then, halve the initial amount nn times. After 3 half-lives, only 18\frac{1}{8} of the original mass remains.

Problem 2:

Complete the decay equation for the Alpha decay of Uranium-238: 92238U→ZATh+24α^{238}_{92}U \rightarrow ^{A}_{Z}Th + ^{4}_{2}\alpha

Solution:

A=238−4=234A = 238 - 4 = 234 Z=92−2=90Z = 92 - 2 = 90 The product is 90234Th\text{The product is } ^{234}_{90}Th

Explanation:

In Alpha decay, the mass number decreases by 4 because 2 protons and 2 neutrons are lost. The atomic number decreases by 2 because 2 protons are lost. This transforms Uranium (Z=92Z=92) into Thorium (Z=90Z=90).

Problem 3:

A radioactive source has an activity of 800 Bq800\text{ Bq}. After 4040 minutes, the activity drops to 50 Bq50\text{ Bq}. What is the half-life of the source?

Solution:

Sequence: 800→1400→2200→3100→450\text{Sequence: } 800 \xrightarrow{1} 400 \xrightarrow{2} 200 \xrightarrow{3} 100 \xrightarrow{4} 50 Number of half-lives (n)=4\text{Number of half-lives } (n) = 4 t1/2=40 minutes4=10 minutest_{1/2} = \frac{40\text{ minutes}}{4} = 10\text{ minutes}

Explanation:

By counting how many times 800800 must be halved to reach 5050, we find that 4 half-lives have occurred. Dividing the total time (4040 mins) by the number of half-lives gives the duration of a single half-life.