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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) AXZ + y → AXZ–1 + a+b                (B) AXZ + 1n0 → A–3XZ–2 + c

        (C) AXZ → AXZ–1 + f                         (D) AXZ + e–1 →AXZ–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).

View this video for more on radioactivity

2.The activity of a sample of radioactive material is R1 at time t1 and R2 at time t(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 = R0e1t1 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/loge 2 T                    (B) λ = 1/loge 2
               (C) λ = 2/T                              (D) λ = loge 2 / T

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

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