Alpha Decay: Definition, Examples, Practice Problems and FAQs

The behavior of radioactive nuclei is different from that of normal nuclei. Let’s suppose you take two weighing pans; you put carbon in one, radioactive uranium in another in suitable conditions. After some time has passed, you would notice that the mass of the carbon remains unchanged, but the mass of uranium has reduced. Now why does this happen? This is due to a phenomenon called radioactive decay or radioactivity; a phenomenon which was intensively experimented upon by scientists like Madam Curie and Henry Becquerel. Due to radioactivity, uranium will slowly decay; when you bring a Geiger-Counter near it (a device used to detect radioactivity), the counter shows deflection, meaning that radiation is emitted from the Uranium. Radioactive decay is of many types. It mainly depends upon the parent nucleus undergoing decay. In this article, we will explore alpha decay in detail. 

Table of Contents

• What is Alpha Decay?
• Radioactivity
• Momentum Conservation in Alpha Decay
• Practice Problems of Alpha Decay
• FAQs of Alpha Decay

What is Alpha(α) Decay?

• An alpha particle is an energetic He - nucleus(₂He⁴) with atomic number 2 and mass number 4. It carries a positive charge of +2e.   
• Alpha particles are emitted when heavy elements like Uranium U-235 undergo radioactive decay. 
• During an decay, the atomic number decreases by 2 and the mass number decreases by 4. Consider an element _ZX^A whose atomic number is Z and mass number is A. Upon undergoing - decay, it transforms to a daughter nucleus Y. The reaction can be represented as, 

         _ZX^A ━━━⟶  _Z-2Y^A-4 + ₂He⁴

Mass of the parent nucleus, m _X = m( _ZX^A) - Zm_e

Mass of the daughter nucleus, mY = m( _Z-2Y^A-4) - (Z - 2)m_e

Mass of the alpha particle mα = m ( ₂He⁴) - 2m_e, m_e - mass of the electron in amu. 

The energy liberated during a nuclear reaction is called its Q-value. The  Q- value of the - decay can be calculated as,  c- velocity of light. 

• >The energy released is available as kinetic energy to the products.
• Example of an  - decay:   ₉₂U²³⁸ decays to Thorium  ₉₀Th²³⁴ emitting an α - particle.

   ₉₂U²³⁸  ━━━⟶  ₉₀Th²³⁴ + ₂He⁴

Radioactivity

• It is the phenomenon through which unstable nuclei become stable by emitting an α particle or β particle. Sometimes, electromagnetic radiations called gamma () rays are produced.
• A β - particle is essentially a positron e⁺ or an electron e^-.
• Radioactivity is a nuclear phenomenon. The phenomenon of α, β or γ radiation depends upon the size of the nucleus undergoing decay and not on other factors like temperature, pressure or other physical conditions.

Momentum Conservation in Alpha Decay

The total momentum is conserved in an alpha decay reaction. Let p_i indicate initial momentum and pf indicate the final momentum. Since the parent nucleus is at rest, p_i=0. 

 _ZX^A  ━━━⟶  _Z-2Y^A-4 + ₂He⁴

Let m indicate the mass of the -particle and my- mass of the daughter nucleus. Let v indicate the recoil velocity of the -nucleus and vy indicate the velocity of the daughter nucleus. Then final momentum p_f = m_αv_α-m_yv_y. 

Since p_f = p_i; m_αv_α - m_yv_y = 0 ⇒ m_αv_{α =} m_yv_y

The α- particle moves in the direction opposite to that of the daughter nucleus. 

Practice Problems of Alpha Decay

Q1) Complete the following reaction

 ₈₈Ra²²⁶  ━━━⟶  ₈₆Rn²²² + ?

Solution) Since there is a decrease in atomic number by 2 and mass number by 4, the above reaction is an - decay.

 ⇒ ₂He⁴ is emitted in the process.

∴  ₈₈Ra²²⁶  ━━━⟶  ₈₆Rn²²² + ₂He⁴

Q2) The total number of  and  particles emitted in the following nuclear reaction is:

 ₉₂U²³⁸  ━━━⟶  ₈₂Pb²¹⁴ is 

1. 6          (b) 8       (c) 7     (d) 10

Solution) b

 ₉₂U²³⁸  ━━━⟶  ₈₂Pb²¹⁴ + 6 ₂He⁴ + 2e^-1

No. of alpha particles = 6

No. of beta particles = 2

Q3) A nucleus having a mass number of 220, initially at rest, emits an α - particle. If the Q - a value of the reaction is 5.5 MeV, then the energy of the α - particle will be 

(a) 4.8 MeV (b)5.4 MeV ( c ) 5.5 MeV (d)6.8 MeV

Solution) b

Let X   ━━━⟶  Y+ ₂He⁴

Given, mass of X = 220

∴ Mass of Y = 220 - 4 = 216

Let K₁ be the kinetic energy of Y and K₂ be the kinetic energy of the  ₂He⁴ nucleus. 

Then K₁ + K₂ = 5.5 MeV -- (i)

Now linear momentum  , hence momentum p1 of Y can be written as  

And linear momentum of particle can be written as p₂ = 

Since momentum is conserved ; p₁=p₂

=  =54 ; K₂=54 = K₁ -- (ii)

Solving (i) and (ii), we get K₂ = 5.4 MeV.

Q4) Alpha particles emitted from a radioactive substance are

(a)Ionized hydrogen nuclei

(b)Doubly ionized helium atom

(c)Uncharged particles having mass equal to proton

(d)negatively charged particles

Solution) b 

Alpha particles are doubly ionized helium atoms.

FAQs of Alpha Decay

Question 1. Heavy nuclei like uranium do not undergo β - decay. Why?

Answer. Heavy nuclei like Uranium undergo α - decay. β - decay occurs in lighter nuclei which have excess protons or neutrons

Question 2. Write the - decay reaction of  ₈₈Ra²²⁶?

Answer.  ₈₈Ra²²⁶  ━━━⟶   ₈₆Rn²²² + ₂He⁴

Question 3. Write the - decay reaction of  ₉₂U²³⁸ to form thorium?

Answer.  ₉₂U²³⁸  ━━━⟶  ₉₀Th²³⁴ + ₂He⁴

Question 4. Write the unit of Q- value?

Answer. The unit of Q- value is MeV.