Radioactive Decay: Definition, Types, Law

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Jasmine Grover

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Radioactive decay (or nuclear decay) is the process caused by radiation to emit the energy of an unstable atomic nucleus. The three most common types of decay are alpha decay (\(\alpha\)-decay), beta decay (\(\beta\)-decay), and gamma decay (\(\gamma\)-decay). Alpha and gamma decay are ruled by the usual electromagnetic and strong forces, where beta decay is governed by the weak force.

Keyterms: Radioactivity, Unit, Ci or curie, Bq (Becquerel), Alpha Decay, Beta Decay, Gamma Decay, Radiation, Electromagnetic force, nuclear decay


What is Radioactive Decay?

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Radioactivity refers to the behaviour projected by the atomic nuclei because of nuclear instability. The nucleus of the atom begins to drain energy in the form of emitted radiation. 

Radioactive Decay
Radioactive Decay

This phenomenon was confirmed through an experiment. A drawer was lined with photographic plates. A small amount of Uranium compound, wrapped in black paper was placed in the drawer. The plates were carefully examined after a while. They showed a clear exposure to radiation. This phenomenon came to be known as Radioactive Decay. 

Radioactive Decay
Radioactive Decay

The presence of an unstable nucleus in the element’s radioisotope leeches out all the energy. This prevents the atom particles from being bounded. The isotopes present are in a state of constant decay in order to stabilise themselves. This causes a major release of energy in the form of radiation. 

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Types of Radioactive Decay

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There are three types i.e Alpha Decay, Beta Decay and Gamma Decay.

Types of Radioactive Decay
Types of Radioactive Decay

Alpha Decay or α-decay

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Alpha decay is a process where a nucleus emits an alpha particle (helium nucleus) into a different atomic nucleus. The following is the formula of alpha decay:

E = (mi − mf − mp) c2

Where,

mi stands for the initial mass of the nucleus

mf stands for the mass of the nucleus after alpha particle emission

mp stands for the mass of the emitted alpha particle

Alpha Decay or α-decay
Alpha Decay or α-decay

Example of Alpha Decay:

238U92234Th90 + 4He2

When 238U92 goes through alpha decay, it transforms into Th23490 and He42 (containing two protons and two neutrons).


Beta Decay or β-decay

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Beta decay is a process where a beta particle (electron/positron) is emitted from an atomic nucleus. The following is the process of beta decay:

234Th90234Pa91 + 0e-1

Beta Decay or β-decay
Beta Decay or β-decay

Gamma Decay or γ-decay

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Just like atoms, the nucleus also has energy levels. When the nucleus' high energy level shifts into a lower energy level, there is an emission of photons having MeV energy, hence, called gamma-ray.

Gamma Decay or γ-decay
Gamma Decay or γ-decay

Radioactive Decay Law

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The radioactive decay law states when the number of radioactive nuclei undergoing the decay (α, β or γ-decay), per unit time, is proportionate to the total number of nuclei in the sample material.

Assuming, N is the number of nuclei in a sample and ΔN is the number of radioactive decays per unit time- Δt, then,

ΔN/Δt ∝ N

Or,

ΔN/Δt = λN

Where,

Λ is the proportionality constant or radioactive decay constant

Also, 

ΔN is the decrease in the total number of nuclei present in the sample

∂N∂t=−λN

After rearranging, it should be

∂NN=−λ∂t

Uniting both will result in:

∫NN0∂NN=−λ∫tt0∂tInN−InN0=−λ(t−t0)

Here,

N0 means the initial number of nuclei present in the sample at a time

Applying t0 which is t=0 in the equation results in:

InNN0=−λt

Leading it to:

Nt=Ne−λt0

The above is an example of exponential type of decay.


Rate of Decay

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From the above results, it is clear to focus on the rates, not on the number. The rate stands for the change per time.

Calculation of the rate of decay:

R=−∂N∂t

Replacing Nt in the equation

Nt=Ne−λt0

Result as

R=−∂N∂t=λNe−λt0R=Re−λt0R0

The decay rate at the time is t=0

Replacing the original equation:

ΔN Δt = λN

Results in 

R = λN


Law Of Radioactive Decay Derivation

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The radioactive decay law is:

ΔN/Δt ∝ N

or

ΔN/Δt = λN

Where,

λ is the radioactive decay constant

The change in the sample along with the number of nuclei is:

dN/dt = −λN (dN/N) = −λdt ∫NN0 dN/N = λ ∫tt0 dt

lnN − lnN0 = −λ (t−t0)

Where,

N0 is the number of radioactive nuclei

t0 is the arbitrary time

Hence, the law of radioactive decay:

ln (N/N0) = −λt N(t) = N0e−λt


Things to Remember

  • Radioactive decay is the energy released in the form of radiation from an unstable atomic nucleus. 
  • Radioactive Decay is of three types: Alpha, Beta and Gamma Radiation
  • Alpha decay is a process where a nucleus emits an alpha particle (helium nucleus) into a different atomic nucleus.
  • Beta decay is a process where a beta particle (electron/positron) is emitted from an atomic nucleus.
  • When the nucleus' high energy level shifts into a lower energy level, there is an emission of photons having MeV energy, called gamma-ray.
  • The radioactive decay law states when the number of radioactive nuclei undergoing the decay (α, β or γ-decay), per unit time, is proportionate to the total number of nuclei in the sample material.

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Sample Questions

Ques: When a nucleus in an atom undergoes radioactive decay, what are the electronic energy levels of the atom? (2 marks)

Ans: When a nucleus in an atom undergoes radioactive decay, the electronic energy levels of an alpha particle is 2 units of positive charge, beta particles are 1 unit positive charge and gamma particles have no charge.

Ques: Mx and My denote the atomic masses of the parent and the daughter nuclei respectively in radioactive decay. The Q-value for a β– decay is Q1 and that for a β+ decay is Q2. If me denotes the mass of an electron, then what will be the value? (3 marks)

Ans: Let’s assume ZXA to be the nucleus, according to ß+ decay will form as:

zXAZ+1YA + +1e0 + v + Q2

Therefore, Q2 = [mN (zXA) - mN (Z-1YA) - me] c2

= [mN (ZXA) + Zme - mn (Z-1YA) - (z-1) me- 2me] c2

= [mN (ZXA) - m (Z-1YA) - 2me] c2

= (Mx -My - 2me) c2

ß- decay will form as:

zXAZ+1AY + -1e0 + v1 + Q1

Therefore, Q1 = [mN (zXA) - mN (z + 1YA) - me] c2 = [mN (zXA) - m (z + 1YA)] c2 = (Mx –My) c2

Ques: Heavy stable nuclei have more neutrons than protons. The reason behind it is? (1 mark)

Ans: Heavy stable nuclei have more neutrons than protons because the electrostatic force between protons is repulsive which results in reducing the nucleus stability.

Ques: Tritium is an isotope of hydrogen whose nucleus Triton contains 2 neutrons and 1 proton. Free neutrons decay into p + ? + ν. If one of the neutrons in Triton decays, it would transform into an He3 nucleus. This does not happen because? (2 marks)

Ans: The nucleus of Tritium (1H3) contains 1 proton and 2 neutrons. A neutron decays as n → p + ? + ν, the nucleus may have 2 protons and 1 neutron, therefore, tritium will transform into 2He3.

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CBSE CLASS XII Related Questions

1.

A parallel plate capacitor made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad s−1.

  1. What is the rms value of the conduction current?
  2. Is the conduction current equal to the displacement current?
  3. Determine the amplitude of B at a point 3.0 cm from the axis between the plates.
A parallel plate capacitor made of circular plates

      2.

      Three capacitors each of capacitance 9 pF are connected in series. 

      (a) What is the total capacitance of the combination? 

      (b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?

          3.
          (a) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60° with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning. 
          (b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)

              4.
              A boy of mass 50 kg is standing at one end of a, boat of length 9 m and mass 400 kg. He runs to the other, end. The distance through which the centre of mass of the boat boy system moves is

                • 0
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                • 3 m

                5.
                A capillary tube of radius r is dipped inside a large vessel of water. The mass of water raised above water level is M. If the radius of capillary is doubled, the mass of water inside capillary will be

                  • 5M
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                  6.

                  An object of size 3.0 cm is placed 14cm in front of a concave lens of focal length 21cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?

                      CBSE CLASS XII Previous Year Papers

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