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Wize University Physics Textbook (Master) > Modern Physics

Alpha Decay

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Radioactive Decays


The most stable situation for nucleus is to be at the minimum possible energy. For this to happen it decays into a more stable nucleus and radiates energy.

  • The original, less stable nucleus is called the parent nucleus (denoted XX here).
  • The resulting, more stable nucleus is called the daughter nucleus (denoted YYhere).


The are 3 different types of radioactive decay: α\bcth\alpha, β\bcth\beta and γ\bcth\gamma decay.

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Wize Concept
The gamma rays are the most powerful and the alpha particles the least powerful.


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Alpha Decay

In an alpha decay, the original nucleus moves to a lower mass nucleus by emitting an alpha particle, 24 α\bcth {_2^4\ \alpha}.

The alpha particle has 2 protons and 2 neutrons, and is identical to a helium nucleus (mass 4.002602 u4.002602 \ u).


Exam Tip
You may see 24He_2^4He instead of 24α_2^4\alpha , they are the same thing!


Both the atomic number ZZ and the mass number AA change:


 Z→Z−2 \boxed {\ Z\to Z-2\ } and  A→A−4 \boxed {\ A\to A-4 \ }

The decay equation is:
 ZAX→ Z−2A−4 Y+ 24 α+energy \boxed{ \ _Z^A X\to \ ^{A-4}_{Z-2} \ Y+\ _2^4\ \alpha+energy \ }

Exam Tip
There is a small recoil for the new nucleus! (conservation of momentum)

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The energy released becomes the kinetic energy of the alpha particles, and can be found using the mass defect:

 E=(mX−mY−mHe) c2 \boxed{ \ E=(m_X-m_Y-m_{He})\ c^2 \ }


Wize Concept
The binding energy per nucleon reaches a maximum for Bismuth with atomic number of 83 and decreases for higher atomic numbers. For atomic numbers higher than 83, the nuclei have the tendency to undergo alpha decay to increase their binding energy per nucleon and become more stable!


Wize Concept
Alpha particles are the heaviest type of radioactive decay. They start off with high energy but rapidly lose it and can be stopped by a sheet of paper!


Example: Alpha Decay

Find the energy emitted when uranium-238 undergoes alpha decay.

Note: you don't have to know the names of the resulting elements unless you're given the periodic table.


Find uranium in the periodic table and write the decay equation, then figure out what element has the atomic number of the daughter nucleus:

 92238U→  90234Th + 24α +energy_{\ 92}^{238}U \to \ _{\ 90}^{234} Th \ + \ _2^4\alpha \ + energy


Find the emitted energy using the mass defect (look up the masses in the periodic table):

E=(mU−mTh−mHe) c2E=(m_U-m_{Th}-m_{He})\ c^2

=(238.0508−234.0436−4.0015)u c2=(238.0508-234.0436-4.0015 )u\ c^2

=0.0057 u c2=0.0057 \ u \ c^2

Let's use u=931.494 MeV/c2u=931.494 \ MeV/c^2 to get:

=0.0057×931.494 MeVc2 c2=0.0057\times931.494\ \dfrac{MeV}{\cancel{c^2}}\ \cancel{c^2}

=5.31 MeV=5.31 \ MeV

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Practice: Chain of Alpha Decays


Thorium-228 decays by emitting an alpha particle, then the daughter nucleus emits an alpha particle again, and so on until four alpha particles are emitted.

a) Write the decay equation for each alpha emission.

b) Write an equation for the total energy emitted.

Note: you don't have to know the names of the resulting elements unless you're given the periodic table.