Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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).

Related resources:

 


TOP Your EXAMS!

Upto 50% Scholarship on Live Classes

Course Features

  • Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution