Nuclei Easy Questions and Answers | Page - 12

Nuclei

A radioactive sample with a half life of 1 month has the label : ‘Activity = 2 micro curies on 1–8–1991. What would be its activity two months earlier ?

1. 1.0 micro curie
1. 0.5 micro curie
1. 4 micro curie
1. 8 micro curie

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In two half lives, the activity becomes one fourth.
Activity on 1–8–91 was 2 micro–curie
∴ Activity before two months,
4 × 2 micro-curie = 8 micro curie

Correct Option: D

In two half lives, the activity becomes one fourth.
Activity on 1–8–91 was 2 micro–curie
∴ Activity before two months,
4 × 2 micro-curie = 8 micro curie

The nucleus ¹¹⁵₄₈Cd, after two successive β– decay will give

1. ¹¹⁵₄₆Pa
1. ¹¹⁴₄₉In
1. ¹¹³₅₀Sn
1. ¹¹⁵₅₀Sn

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Two successive β decay increases the atomic number by 2. Therefore, (d) is correct.

Correct Option: D

Two successive β decay increases the atomic number by 2. Therefore, (d) is correct.

In the nuclear decay given below:
A X → A Y → A - 4 B* → A - 4 B,
Z Z + 1 Z - 1 Z - 1

the particles emitted in the sequence are

1. γ, β, α
1. β, γ, α
1. α, β, γ
1. β, α, γ

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A X → A Y:β , A Y → A - 4 B*:α
Z Z + 1 Z + 1 Z - 1

A - 4 B* → A - 4 B:γ
Z - 1 Z - 1

(β, α, γ) (∴ β = 0 e, α = 4 He,
1 2
mass number and charge number of a nucleus remains unchanged during γ decay)

Correct Option: D

A X → A Y:β , A Y → A - 4 B*:α
Z Z + 1 Z + 1 Z - 1

A - 4 B* → A - 4 B:γ
Z - 1 Z - 1

(β, α, γ) (∴ β = 0 e, α = 4 He,
1 2
mass number and charge number of a nucleus remains unchanged during γ decay)

Two radioactive materials X₁ and X₂ have decay constants 5λ and λ respectively. If initially they have the same number of nuclei,
then the ratio of the number of nuclei of X₁ to that of X₂ will be 1/e after a time

1. λ
1. 1 λ
  2
1. 1
  4λ
1. e
  λ

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Let the required time be t. Then
N₁ = N₀e^-λ₁t ; N₂N₀e^-λ₂t
Where
N₁ = number of nuclei of X₁1 after time t
N₂ = number of nuclei of X₂ after time t
N₀ = initial number of nuclei of X₁ and X₂ each.
Now, N₁ = N₀e^-λ₁t
N₂ N₀e^-λ₂t

Here N₁ = 1
N₂ e

λ₁ = 5λ ; λ₂ = λ
∴ 1 = e^-5λt
e e^-λt

⇒ e^-1 = e^-4λt ⇒ 44λt = 1
∴ t = 1
4λ

Correct Option: C

Let the required time be t. Then
N₁ = N₀e^-λ₁t ; N₂N₀e^-λ₂t
Where
N₁ = number of nuclei of X₁1 after time t
N₂ = number of nuclei of X₂ after time t
N₀ = initial number of nuclei of X₁ and X₂ each.
Now, N₁ = N₀e^-λ₁t
N₂ N₀e^-λ₂t

Here N₁ = 1
N₂ e

λ₁ = 5λ ; λ₂ = λ
∴ 1 = e^-5λt
e e^-λt

⇒ e^-1 = e^-4λt ⇒ 44λt = 1
∴ t = 1
4λ

Two radioactive substances A and B have decay constants 5 λ and λ respectively. at t = 0 they have same number of nulei. The ratio of number of nuclei of A to those of B will be (1/e)² after a time interval

1. 4 λ
1. 2λ
1. 1/2λ
1. 1/4λ

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λ_A = 5λ and λ_B = λ
At t = 0, (N₀)_A = (N₀)_B
Given,
N_A = 1 ²
N_B e

According to radioactive decay,
N = e^λt
N₀

N_A = e^λ_At ....(i)
(N₀)_A

N_B = e^λ_Bt ....(ii)
(N₀)_B

From (1) and (2),
N_A = e^{-(5λ - λ)t}
N_B

⇒ 1 ² = e^-4λt 1 ^4λt
e e

⇒ 4λt = 2
∴ t = 1 .
2λ

Correct Option: C

λ_A = 5λ and λ_B = λ
At t = 0, (N₀)_A = (N₀)_B
Given,
N_A = 1 ²
N_B e

According to radioactive decay,
N = e^λt
N₀

N_A = e^λ_At ....(i)
(N₀)_A

N_B = e^λ_Bt ....(ii)
(N₀)_B

From (1) and (2),
N_A = e^{-(5λ - λ)t}
N_B

⇒ 1 ² = e^-4λt 1 ^4λt
e e

⇒ 4λt = 2
∴ t = 1 .
2λ