40 K isotope of potassium has a half life of 1.37 × 109 years and decays to an isotope of argon which is stable. In a particular sample of moon rock, the ratio of potassium atoms to argon atoms was found to be 1: 7. The age of the rock, assuming that originally there was no argon present, is

Question

40 K isotope of potassium has a half life of 1.37 × 109 years and decays to an isotope of argon which is stable. In a particular sample of moon rock, the ratio of potassium atoms to argon atoms was found to be 1: 7. The age of the rock, assuming that originally there was no argon present, is (A) 4.11 × 109 year (B) 2.74 × 109 year (C) 5.48 × 109 year (D) 1.37 × 109 year

Tardigrade Admin · Accepted Answer

T 1 / 2=1.37 × 109 year Ratio - Potassium: Argon =1: 7 Amount of Potassium left =(1/8) ((1/2))n=(1/8) ∴ n or number of half lives =3

$4.11 \times 1 0^{9}$ year

$2.74 \times 1 0^{9}$ year

$5.48 \times 1 0^{9}$ year

$1.37 \times 1 0^{9}$ year

$T_{1/2} = 1.37 \times 1 0^{9}$ year
Ratio $-$ Potassium : Argon $= 1 : 7$
Amount of Potassium left $= \frac{1}{8}$
$(\frac{1}{2})^{n} = \frac{1}{8}$
$∴ n$ or number of half lives $= 3$

Q. 40K isotope of potassium has a half life of 1.37×109 years and decays to an isotope of argon which is stable. In a particular sample of moon rock, the ratio of potassium atoms to argon atoms was found to be 1:7. The age of the rock, assuming that originally there was no argon present, is

4.11×109 year

2.74×109 year

5.48×109 year

1.37×109 year

T1/2​=1.37×109 year Ratio − Potassium : Argon =1:7 Amount of Potassium left =81​ (21​)n=81​ ∴n or number of half lives =3

$4.11 \times 1 0^{9}$ year

$2.74 \times 1 0^{9}$ year

$5.48 \times 1 0^{9}$ year

$1.37 \times 1 0^{9}$ year

$T_{1/2} = 1.37 \times 1 0^{9}$ year
Ratio $-$ Potassium : Argon $= 1 : 7$
Amount of Potassium left $= \frac{1}{8}$
$(\frac{1}{2})^{n} = \frac{1}{8}$
$∴ n$ or number of half lives $= 3$