At radioactive equilibrium- the ratio between two atoms of radioactive elements A and B is 3-1- 109-1- If the half-life period of A is 2- 1010- then the half-life if B is-

At radioactive equilibrium, the ratio between the number of the atoms of two radioactive elements (X) and (Y) was found to be $3.2 \times 10^{9}$ : 1 respectively. If half-life of the element (X) is $1.6 \times 10^{10}$ years, then half-life of the element (Y) would be

Q. The ratio of the atoms of two elements

A

and

B

at radioactive equilibrium is

5.0 \times 10^{5} : 1

respectively. Calculate half life of

B

if half life of

A

245

days.

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At radioactive equilibrium, the ratio between two atoms of radioactive elements A and B is 3.1×109:1. If the half-life period of A is 2×1010, then the half-life if B is:

The correct option is D 6.45 yrsAt radioactive equilibrium,A⟶BNANB=KBKA=t1/2At1/2B∴3.1×1091=2×1010t1/2B t1/2B=6.45 yrs. Hence, option D is correct.

At radioactive equilibrium, the ratio between two atoms of radioactive elements A and B is $3.1 \times 10^{9} : 1$ . If the half-life period of A is $2 \times 10^{10}$ , then the half-life if B is:

The correct option is D 6.45 yrs
At radioactive equilibrium,
$A ⟶ B$

$\frac{N_{A}}{N_{B}} = \frac{K_{B}}{K_{A}} = \frac{t_{1 / 2 A}}{t_{1 / 2 B}}$

$∴ \frac{3.1 \times 10^{9}}{1} = \frac{2 \times 10^{10}}{t_{1 / 2 B}}$

$t_{1 / 2 B} = 6.45 y r s$ .

Hence, option D is correct.