Solved Examples on Radioactivity Calculate Decay Constant

Radioactivity is one of the most important topics of Modern Physics. It is important to have a thorough knowledge of all the three rays i.e. alpha, beta and gamma rays. Solving problems of gamma decay in Modern Physics is not a very tough task but requires clarity of concepts. We discuss here some questions on calculation of decaying constant.

1.Which of the following process represents a -decay?

        (A) ^AX_Z + y → ^AX_Z–1 + a+b                (B) ^AX_Z + ¹n₀ → ^A–3X_Z–2 + c

        (C) ^AX_Z → ^AX_Z–1 + f                         (D) ^AX_Z + e_–1 →^AX_Z–1 + g

Solution: We know that in gamma decay only the quantum states of the nucleons change. It does not bring about any change in the proton number (Z) or the neutron number (A-Z). Hence, the correct option is (C).

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2.The activity of a sample of radioactive material is R₁ at time t₁ and R₂ at time t₂ (t₂ > t₁). If mean life of the radioactive sample is T, then:
(A) R₁t₁ = R₂t₂                           (B) R₁ – R₂ / t₂ – t₁ = constant 
(B)  R₂ = R₁exp (t₁ – t₂ / T)        (D) R₂ = R₁exp (t₁ / Tt₂)

Solution: Let us assume that R₀ is the initial activity of the sample,
Then R₁ = R₀e^1t₁ and R₂ = R₀e^–1t₂ 
 
Hence, the correct option is (C).
 
3. When ₁₅P³⁰ decays to become ₁₄Si³⁰, the particle released is
        (A) Electron                            (B) α-particle
        (C) Neutron                            (D) positron

Sol: The reaction which occurs when ₁₅P³⁰ decays to become ₁₄Si³⁰  is
     ₁₅P³⁰ → + ₁e⁰ + ₁₄Si³⁰ 
Hence, it is clear that the particle that is released is (D) positron.
 
4. Among electron, proton, neutron and α-particle the maximum penetration capacity is for
           (A) Electron                            (B) proton 
           (C) Neutron                            (D) α-particle

Sol: The maximum penetration capacity is for Neutron. So (C) is the correct option. 
 
5. Plutonium decays with a half-life of 24000 years. If plutonium is stored for 72000 years, the fraction of it that remains is
        (A) 1/2                                      (B) 1/3
        (C) 1/4                                      (D) 1/8

Solution: The half-life of Plutonium is given to be 24000 years. It is stored for a period of 72000 years.
So, the number of half-life periods = 72000/24000 = 3
The fraction that remains is given by 1/2³ = 1/8. This gives (D) as the correct option.
 
6. The decay constant (λ) and the half-life (T) of a radioactive isotope are related as:
               (A) λ = 1/log_e 2 T                    (B) λ = 1/log_e 2
               (C) λ = 2/T                              (D) λ = log_e 2 / T

Solution: The relation between the half-life and the decay constant is given by
Half-life T = 0.693 / λ 
=> λ = log_e 2 / T. so the answer is (D).

Related resources:

• Click here for the Detailed Syllabus of IIT JEE Physics .

• Look into the Sample Papers with Solutions to get a hint of the kinds of questions asked in the exam.

• You can get the knowledge of Recommended Books of Physics here.