The initial activity of a certain radioactive isotope was measured as 16000 counts min -1. Given that the only activity measured was due to this isotope and that its activity after 12 h was 2100 counts min -1, its half-life, in hours, is nearest to [Given log e(7.2)=2 ]

Question

The initial activity of a certain radioactive isotope was measured as 16000 counts min -1. Given that the only activity measured was due to this isotope and that its activity after 12 h was 2100 counts min -1, its half-life, in hours, is nearest to [Given log e(7.2)=2 ] (A) 9.0 (B) 6.0 (C) 4.0 (D) 3.0

Tardigrade Admin · Accepted Answer

A=A0 e-λ t ; 2100=16000 e-12 λ ⇒ e12 λ=7.6 ⇒ 12 λ= log e 7.6=2 ⇒ λ=(2/12)=(1/6) ∴ T=(0.6931 × 6/1)=4

Q. The initial activity of a certain radioactive isotope was measured as $16000$ counts $min^{- 1}$ . Given that the only activity measured was due to this isotope and that its activity after $12 h$ was $2100$ counts $mi n^{- 1}$ , its half-life, in hours, is nearest to [Given $lo g_{e} (7.2) = 2$ ]

9.0

6.0

4.0

3.0

$A = A_{0} e^{- λ t}; 2100 = 16000 e^{- 12 λ}$
$\Rightarrow e^{12 λ} = 7.6$
$\Rightarrow 12 λ = lo g_{e} 7.6 = 2$
$\Rightarrow λ = \frac{2}{12} = \frac{1}{6}$
$∴ T = \frac{0.6931 \times 6}{1} = 4$

Q. The initial activity of a certain radioactive isotope was measured as 16000 counts min−1. Given that the only activity measured was due to this isotope and that its activity after 12h was 2100 counts min−1, its half-life, in hours, is nearest to [Given loge​(7.2)=2 ]

9.0

6.0

4.0

3.0

A=A0​e−λt;2100=16000e−12λ ⇒e12λ=7.6 ⇒12λ=loge​7.6=2 ⇒λ=122​=61​ ∴T=10.6931×6​=4

$A = A_{0} e^{- λ t}; 2100 = 16000 e^{- 12 λ}$
$\Rightarrow e^{12 λ} = 7.6$
$\Rightarrow 12 λ = lo g_{e} 7.6 = 2$
$\Rightarrow λ = \frac{2}{12} = \frac{1}{6}$
$∴ T = \frac{0.6931 \times 6}{1} = 4$