Polonium has a half-life of 140 days. If we take 20 g of polonium initially then the amount of it that remains after 280 days is

Question

Polonium has a half-life of 140 days. If we take 20 g of polonium initially then the amount of it that remains after 280 days is (A) 2.5 g (B) 5 g (C) 10 g (D) 15 g

Tardigrade Admin · Accepted Answer

After n-half lines, quantity of radiatioactive substance left is N=N0((1/2))t / T1 / 2 Given, T1 / 2=140 days, t=28 days, N0=20 g ∴ N=20((1/2))(280/140) =(20/4)=5 g

Q. Polonium has a half-life of $140$ days. If we take $20 g$ of polonium initially then the amount of it that remains after $280$ days is

2.5 g

5 g

10 g

15 g

After n-half lines, quantity of radiatioactive substance left is
$N = N_{0} (\frac{1}{2})^{t / T_{1/2}}$
Given, $T_{1/2} = 140$ days,
$t = 28$ days, $N_{0} = 20 g$
$∴ N = 20 (\frac{1}{2})^{\frac{280}{140}}$
$= \frac{20}{4} = 5 g$

Q. Polonium has a half-life of 140 days. If we take 20g of polonium initially then the amount of it that remains after 280 days is

2.5 g

5 g

10 g

15 g

After n-half lines, quantity of radiatioactive substance left is N=N0​(21​)t/T1/2​ Given, T1/2​=140 days, t=28 days, N0​=20g ∴N=20(21​)140280​ =420​=5g

Q. Polonium has a half-life of $140$ days. If we take $20 g$ of polonium initially then the amount of it that remains after $280$ days is

After n-half lines, quantity of radiatioactive substance left is
$N = N_{0} (\frac{1}{2})^{t / T_{1/2}}$
Given, $T_{1/2} = 140$ days,
$t = 28$ days, $N_{0} = 20 g$
$∴ N = 20 (\frac{1}{2})^{\frac{280}{140}}$
$= \frac{20}{4} = 5 g$