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Physics - Radioactivity

Grade 10ICSE

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

🔑Concepts

Radioactivity is a nuclear phenomenon; it is a spontaneous and random process by which an unstable nucleus emits radiations like α\alpha, β\beta, and γ\gamma to reach a stable state.

The composition of an atom is represented as ZAX_{Z}^{A}X, where AA is the mass number (protons + neutrons) and ZZ is the atomic number (protons).

α\alpha-particles are helium nuclei (24He_{2}^{4}He) with a charge of +2e+2e and mass of 4u4\,u. They have the highest ionizing power but the least penetrating power.

β\beta-particles are fast-moving electrons (10e_{-1}^{0}e) emitted from the nucleus when a neutron converts into a proton: 01n11p+10e_{0}^{1}n \rightarrow _{1}^{1}p + _{-1}^{0}e.

γ\gamma-radiations are electromagnetic waves of very short wavelength (approx 1013m10^{-13}\,m). They have the highest penetrating power and travel at the speed of light (c3×108m/sc \approx 3 \times 10^8\,m/s).

In α\alpha-decay, the mass number AA decreases by 44 and the atomic number ZZ decreases by 22.

In β\beta-decay, the mass number AA remains unchanged and the atomic number ZZ increases by 11.

Isotopes are atoms of the same element having the same ZZ but different AA. Isobars have the same AA but different ZZ. Isotones have the same number of neutrons (AZ)(A - Z).

📐Formulae

Alpha Decay: ZAXZ2A4Y+24He\text{Alpha Decay: } _{Z}^{A}X \rightarrow _{Z-2}^{A-4}Y + _{2}^{4}He

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

Gamma Emission: ZAXZAX+γ\text{Gamma Emission: } _{Z}^{A}X^* \rightarrow _{Z}^{A}X + \gamma

Neutron Conversion: 01n11p+10e\text{Neutron Conversion: } _{0}^{1}n \rightarrow _{1}^{1}p + _{-1}^{0}e

Mass-Energy Equivalence: E=Δmc2\text{Mass-Energy Equivalence: } E = \Delta m c^2

💡Examples

Problem 1:

A radioactive nucleus 92238U_{92}^{238}U emits an α\alpha-particle to form a nucleus XX. This nucleus XX then emits a β\beta-particle to form a nucleus YY. Find the atomic number and mass number of YY.

Solution:

Step 1: α\alpha-decay of Uranium: 92238U9222384X+24He    90234X_{92}^{238}U \rightarrow _{92-2}^{238-4}X + _{2}^{4}He \implies _{90}^{234}X. Step 2: β\beta-decay of XX: 90234X90+1234Y+10e    91234Y_{90}^{234}X \rightarrow _{90+1}^{234}Y + _{-1}^{0}e \implies _{91}^{234}Y.

Explanation:

In α\alpha-decay, the mass number decreases by 44 (2384=234238 - 4 = 234) and the atomic number decreases by 22 (922=9092 - 2 = 90). In β\beta-decay, the mass number remains the same (234234) and the atomic number increases by 11 (90+1=9190 + 1 = 91).

Problem 2:

Explain why the mass number of a nucleus does not change during β\beta-decay even though a particle is emitted.

Solution:

During β\beta-decay, a neutron inside the nucleus converts into a proton and an electron (01n11p+10e_{0}^{1}n \rightarrow _{1}^{1}p + _{-1}^{0}e).

Explanation:

Since the mass number AA is the total sum of protons and neutrons, the loss of one neutron is exactly balanced by the gain of one proton. Thus, AA remains constant while ZZ increases by 11.