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# Solved Examples on Half-time and Average-Life

The half-life of a radioactive substance represents the time required for a radioactive isotope to decrease by a factor of two. It may be written as

Nt = N0/2.

The half- life does not specify the duration of time period for which a radioactive substance will remain radioactive but it just tells the time for its radioactivity to halve. Some elements have a relatively short half-life. These are the elements which are used for medical purposes as they do not remain radioactive for long and hence result in a low radiation dose. Q1.The decay constant of a radioactive sample is λ. The half-time and the average-life of the sample will be respectively

(A) 1/λ and (In2)/ λ                     (B) (In2)/ λ and 1/ λ

(C) λ (In2) and 1/λ                      (D) λ/ (In2) and 1/ λ

Solution: We are required to calculate the half-life and average life.

We know that the average life is 1/λ. Now, this deletes the first option. But now we know that the half-life is 0.693/ λ. We calculate the value of ln 2.

ln 2 = 0.301 × 2.303 = 0.693. so the correct option is (B).

Q2. In the uranium radioactive series, the initial nucleus is 92U238 and the final nucleus is 82Pb206. When the uranium nucleus decays to lead, the number of α-particle emitted will be

(A) 1                              (B) 2

(C) 4                              (D) 8

Solution: We need to consider the radioactive series of uranium. The number of α particles emitted will be given by

an = A1 – A2 / 4 = 238 – 206 / 4 = 32 / 4 = 8. So (D) is the correct option.

Q3. The activity of a radioactive sample is 1.6 curie, and its half-life is 2.5 days. Its activity after 10 days will be:

(A) 0.8 curie                    (B) 0.4 curie

(C) 0.1 curie                    (D) 0.16 curie

Solution: The half-life of the radioactive substance is 2.5 days.

So, n(10/2.5) = 4

A/A0 = (1/2)n

=> A = 1.6 (1/2)4

So, A = 0.1 Curie.  Hence, the correct answer is (C).

Q4. In a mean life of a radioactive sample:

(A)    About 1/3 of substance disintegrates

(B)    About 2/3 of the substance disintegrates

(C)    About 90% of the substance disintegrates

(D)    Almost all the substance disintegrates.

Solution: We know the formula

t = 1/λ log N0 / N

Given that t = (1/λ)

1/λ = 1/λ log N0/N

=> log N0/N = 1

N0/N = 2.7

=> N = N0/2.7 = 0.37 N0 ~ 1/3 N0. So (B) is the correct option.

Q5. Concept of neutrino was given by

(A) Sommerfield               (B) Pauli

Solution: Pauli was the first one to have initiated the concept of a neutrino. Hence, (B) is the correct answer.

Q6. Half-life of an element is 30 days. How much part will remain after 90 days?

(A) 1/4th part                 (B) 1/16 part

(C) 1/8th part                 (D) 1/3rd part

Solution: The half-life is given to be 30 days. Then we know the relation

N/N0 = 1/2(1/T) = 1/2(90/30) = 1/8. This gives (C) option as the correct answer.

Watch this Video for more reference

7. In the radioactive decay 92X234  87Y222 number of emitted α and β particles emitted are

(A) 3, 3                           (B) 3, 1

(C) 5, 3                           (D) 3, 5

Solution: The given reaction is 92X234 → 87Y222.

We know that the number of α and β emissions are given by

nα = A1 – A2 = 234 – 222 / 4 = 3

nβ = Z2 + 2na – Z1 = 87 + 6 – 92 = 1

Hence, (B) is the correct option.

8. After five half-lives what will be the fraction of initial substance remaining undecayed

(A) (1/2)10                            (B)    (1/2)5

(C) (1/2)4                              (D)    1/2

Solution: It is quite obvious that after five half-lives, the fraction of the initial substance remaining is given by

N/N0 = (1/2)5, which given (B) as the correct option.

9. If half-life of a substance is 3.8 days & its initial quantity is 10.38 gm, then quantity of substance reaming after 19 days will be:

(A)    0.151 gm                   (B)    0.32 gm

(C)    1.51 gm                     (D)    0.16 gm

Solution: The half-life of the substance is given to be 3.8 days and the initial quantity is given to be 10.38 gm.

The number of half-lives =19/ 3.8 = 5

Quantity remaining after n half lives  = (1/2)n No. = (1/2)5

= 10.38/32 = 0.32 gm which is the same as (B).

It is important to have expertise on these topics to remain competitive in the JEE. Questions are often framed on calculating average life of uranium or average life of a radioactive sample. askIITians provides comprehensive study material containing numerous solved examples on these topics.