In a sample of radioactive -material, what percentage of the initial number of active nuclei will decay during one mean life?

Question

In a sample of radioactive -material, what percentage of the initial number of active nuclei will decay during one mean life? (A) 37% (B) 50% (C) 63% (D) 69.3%

Tardigrade Admin · Accepted Answer

From Rutherford-Soddy law N=N0 e-λ τ ∴ Using τ=(1/λ), we have N=N0 e-1=(N0/e) Percentage decayed =(N0-N/N0) × 100 =((N0-N0 / e)/N0) × 100 =63 %

Q. In a sample of radioactive -material, what percentage of the initial number of active nuclei will decay during one mean life?

37%

50%

63%

69.3%

From Rutherford-Soddy law
$N = N_{0} e^{- λ τ}$
$∴$ Using $τ = \frac{1}{λ}$ , we have
$N = N_{0} e^{- 1} = \frac{N _{0}}{e}$
Percentage decayed $= \frac{N _{0} - N}{N _{0}} \times 100$
$= \frac{( N _{0} - N _{0} / e )}{N _{0}} \times 100$
$= 63%$

Q. In a sample of radioactive -material, what percentage of the initial number of active nuclei will decay during one mean life?

37%

50%

63%

69.3%

From Rutherford-Soddy law N=N0​e−λτ ∴ Using τ=λ1​, we have N=N0​e−1=eN0​​ Percentage decayed =N0​N0​−N​×100 =N0​(N0​−N0​/e)​×100 =63%

From Rutherford-Soddy law
$N = N_{0} e^{- λ τ}$
$∴$ Using $τ = \frac{1}{λ}$ , we have
$N = N_{0} e^{- 1} = \frac{N _{0}}{e}$
Percentage decayed $= \frac{N _{0} - N}{N _{0}} \times 100$
$= \frac{( N _{0} - N _{0} / e )}{N _{0}} \times 100$
$= 63%$