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Physics: Nuclear Physics - Alpha, Beta, and Gamma Radiation

Grade 8IB

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

🔑Concepts

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Radioactivity is the process by which an unstable atomic nucleus loses energy by radiation. The three main types are Alpha (α\alpha), Beta (β\beta), and Gamma (γ\gamma).

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Atomic notation is written as ZAX^{A}_{Z}X, where AA is the mass number (protons + neutrons) and ZZ is the atomic number (protons).

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Alpha particles (24α^{4}_{2}\alpha or 24He^{4}_{2}He) consist of two protons and two neutrons. They are highly ionizing but have low penetration power, being stopped by a sheet of paper.

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Beta particles (−10β^{0}_{-1}\beta or −10e^{0}_{-1}e) are high-energy electrons emitted when a neutron in the nucleus turns into a proton. They have moderate ionizing and penetration power, being stopped by a few millimeters of aluminum.

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Gamma radiation (00γ^{0}_{0}\gamma) consists of high-frequency electromagnetic waves. They have very low ionizing power but high penetration power, requiring several centimeters of lead or meters of concrete to be stopped.

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During Alpha decay, the mass number decreases by 4 and the atomic number decreases by 2. During Beta decay, the mass number remains the same and the atomic number increases by 1.

📐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+−10e\text{Beta Decay: } ^{A}_{Z}X \rightarrow ^{A}_{Z+1}Y + ^{0}_{-1}e

Gamma Emission: ZAX∗→ZAX+00γ\text{Gamma Emission: } ^{A}_{Z}X^{*} \rightarrow ^{A}_{Z}X + ^{0}_{0}\gamma

💡Examples

Problem 1:

An isotope of Uranium-238 (92238U^{238}_{92}U) undergoes alpha decay. Determine the resulting daughter nucleus.

Solution:

92238U→90234Th+24α^{238}_{92}U \rightarrow ^{234}_{90}Th + ^{4}_{2}\alpha

Explanation:

In alpha decay, the mass number decreases by 4 (238−4=234238 - 4 = 234) and the atomic number decreases by 2 (92−2=9092 - 2 = 90). The element with atomic number 90 is Thorium (ThTh).

Problem 2:

Carbon-14 (614C^{14}_{6}C) is unstable and undergoes beta decay. Write the balanced nuclear equation.

Solution:

614C→714N+−10e^{14}_{6}C \rightarrow ^{14}_{7}N + ^{0}_{-1}e

Explanation:

In beta decay, a neutron changes into a proton and an electron. The mass number remains 1414, while the atomic number increases from 66 to 77. The element with atomic number 7 is Nitrogen (NN).

Problem 3:

Calculate the number of neutrons in a Radium-226 nucleus (88226Ra^{226}_{88}Ra).

Solution:

N=A−ZN = A - Z N=226−88=138N = 226 - 88 = 138

Explanation:

The number of neutrons is found by subtracting the atomic number (ZZ) from the mass number (AA).