Nuclei Easy Questions and Answers | Page - 18

Nuclei

A radioactive element has half life period 800 years. After 6400 years what amount will remain?

1. 1/2
1. 1/16
1. 1/8
1. 1/256

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No. of half lives,
n = t = 6400 = 8
T 800

N = 1 ⁸
N₀ 2

= 1
256

Correct Option: D

No. of half lives,
n = t = 6400 = 8
T 800

N = 1 ⁸
N₀ 2

= 1
256

The half life of a radioactive nucleus is 50 days. The time interval (t₂ – t₁ ) between the time t₂ when 2/3 of it has decayed and the time t₁ when 1/3 of it had decayed is :

1. 30 days
1. 50 days
1. 60 days
1. 15 days

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N₁ = N₁ e^–λt
N₁ = 1 N₀
3

N₀ = N₀e^{- λt₂}
3

1 = e^{- λt₂} ....(i)
3

N₂ = 2 N₀
3

2 N₀ = N₀e^{- λt₁}
3

2 = e^{- λt₁}
3

Dividing equation (i) by equation (ii)
1 = e^{- λ(t₂ - t₁)}
2

λ(t₂ - t₁) In 2
t₂ - t₁ = In 2 = T_1/2 = 50 days
λ

Correct Option: B

N₁ = N₁ e^–λt
N₁ = 1 N₀
3

N₀ = N₀e^{- λt₂}
3

1 = e^{- λt₂} ....(i)
3

N₂ = 2 N₀
3

2 N₀ = N₀e^{- λt₁}
3

2 = e^{- λt₁}
3

Dividing equation (i) by equation (ii)
1 = e^{- λ(t₂ - t₁)}
2

λ(t₂ - t₁) In 2
t₂ - t₁ = In 2 = T_1/2 = 50 days
λ

The half life of a radioactive isotope 'X' is 50 years. It decays to another element 'Y' which is stable. The two elements 'X' and 'Y' were found to be in the ratio of 1 : 15 in a sample of a given rock. The age of the rock was estimated to be

1. 150 years
1. 200 years
1. 250 years
1. 100 years

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Let number of atoms in X = N_x
Number of atoms in Y = N_y
By question
N_x = 1
N₀ 15

∴ Part of N_x = 1 (N_x + N_y)
16

= 1 (N_x + N_y)
2⁴

So, total 4 half lives are passed, so, age of rock is 4 × 50 = 200 years

Correct Option: B

Let number of atoms in X = N_x
Number of atoms in Y = N_y
By question
N_x = 1
N₀ 15

∴ Part of N_x = 1 (N_x + N_y)
16

= 1 (N_x + N_y)
2⁴

So, total 4 half lives are passed, so, age of rock is 4 × 50 = 200 years

A nucleus _nX^m emits one α-particle and two β-particles. The resulting nucleus is

1. m - 6 Z
  n - 4
1. m - 6 Z
  n
1. m - 4 X
  n
1. m - 4 Y
  n - 2

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When ^m_nX emits one α-particle then its atomic mass decreases by 4 units and atomic number by 2. Therefore, the new nucleus becomes ^m-4_n-2Y . But as it emits two β– particles, its atomic number increases by 2. Thus the resulting nucleus is ^m-4_nX.

Correct Option: C

When ^m_nX emits one α-particle then its atomic mass decreases by 4 units and atomic number by 2. Therefore, the new nucleus becomes ^m-4_n-2Y . But as it emits two β– particles, its atomic number increases by 2. Thus the resulting nucleus is ^m-4_nX.

Two radioactive nuclei P and Q, in a given sample decay into a stable nucleolus R. At time t = 0, number of P species are 4 N₀ and that of Q are N₀ . Half-life of P (for conversion to R) is 1 minute where as that of Q is 2 minutes. Initially there are no nuclei of R present in the sample. When number of nuclei of P and Q are equal, the number of nuclei of R present in the sample would be

1. 3N₀
1. 9N₀
  2
1. 5N₀
  2
1. 2N₀

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Initially P → 4N₀
Q → N₀
Half life T_P = 1 min.
T_Q = 2 min.
Let after time t number of nuclei of P and Q are equal, that is
4N₀ = N₀
2^t/1 2^t/2

⇒ 4N = 1
2^t/1 2^t/2

⇒ 2^t/1 = 4.2^t/2
2².2^t/2 = 2^{(2 + t/2)}
⇒ t = 2 + t
1 2

⇒ t = 2
2

⇒ t = 4 min
N_P = 4N₀ = N₀
2^4/1 4

at t = 4 min.
N₀ = N₀ = N₀
4 4

or population of R
4N₀ - N₀ + N₀ - N₀
4 4

= 9N₀
2

Correct Option: B

Initially P → 4N₀
Q → N₀
Half life T_P = 1 min.
T_Q = 2 min.
Let after time t number of nuclei of P and Q are equal, that is
4N₀ = N₀
2^t/1 2^t/2

⇒ 4N = 1
2^t/1 2^t/2

⇒ 2^t/1 = 4.2^t/2
2².2^t/2 = 2^{(2 + t/2)}
⇒ t = 2 + t
1 2

⇒ t = 2
2

⇒ t = 4 min
N_P = 4N₀ = N₀
2^4/1 4

at t = 4 min.
N₀ = N₀ = N₀
4 4

or population of R
4N₀ - N₀ + N₀ - N₀
4 4

= 9N₀
2