SATHEE: Atoms And Nuclei Question 409

Atoms And Nuclei Question 409

Question: A radioactive element decays by p emission. A detector records n beta particles in 2 second and in next 2 seconds it records $0.75 n$ beta particles. Find mean life corrected to nearest whole number. Given $ℓ n 2 = 0.6931$ and $ℓ n 3 = 1.0986 .$ .

Options:

A) $6.9 s e c$

B) $9.9 s e c$

C) $10.1 s e c$

D) $12.2 s e c$

Show Answer

Answer:

Correct Answer: A

Solution:

We know that, $N = N_{0} e^{- λ t}$

After 2 second, $N_{1} = N_{0} e^{- λ \times 2}$ After $(2 + 2)$ second, $N_{2} = N_{0} e^{- λ \times 4}$ According to given conditions, $N_{0} - N_{1} = n$ or $N_{0} - N_{0} e^{- 2 λ} = n$ …(i) and

$N_{0} e^{- 2 λ} - N_{0} e^{- 4 λ} = 0.75, n$ … (ii)

After solving above equations, we get $λ = 0.145, s$ Mean life $T = 1 / λ = 6.9 s$ .

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Atoms And Nuclei Question 409

Question: A radioactive element decays by p emission. A detector records n beta particles in 2 second and in next 2 seconds it records 0.75n beta particles. Find mean life corrected to nearest whole number. Given ℓn2=0.6931 and ℓn3=1.0986. .

Options:

Answer:

Solution:

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Question: A radioactive element decays by p emission. A detector records n beta particles in 2 second and in next 2 seconds it records $0.75 n$ beta particles. Find mean life corrected to nearest whole number. Given $ℓ n 2 = 0.6931$ and $ℓ n 3 = 1.0986 .$ .