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III Long answer questions · Q7

Q.Discuss the alpha decay process with an example.

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Step 1. The decay equation. In alpha decay, an unstable nucleus emits a 24He^{4}_{2}He nucleus (two protons, two neutrons); its atomic number decreases by 2 and mass number decreases by 4:

ZAX→Z−2A−4Y+24He{}^{A}_{Z}X\to{}^{A-4}_{Z-2}Y+{}^{4}_{2}He

Step 2. Concrete example. Uranium-238 decays to thorium-234 with the emission of an alpha particle: 92238U→90234Th+24He^{238}_{92}U\to{}^{234}_{90}Th+{}^{4}_{2}He.

Step 3. Why a whole 24He^4_2He, not four separate nucleons. Because 24He^4_2He is itself tightly bound, its mass is much less than four separate free nucleons, so this channel gives a positive disintegration energy QQ; emitting four unbound nucleons instead would give Q<0Q<0, forbidden by energy conservation.

Step 4. Disintegration energy. The mass lost, Δm=mX−mY−mα\Delta m=m_X-m_Y-m_\alpha, converts to released kinetic energy Q=(Δm)c2Q=(\Delta m)c^2, shared between the daughter nucleus YY and the alpha particle. …

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