The half-life of a radioactive nuclide is 100 hours. The fraction of original activity that will remain after 150 hours would be :

Question

The half-life of a radioactive nuclide is 100 hours. The fraction of original activity that will remain after 150 hours would be : (A) 1 / 2 (B) (1/2 √2) (C) (2/3) (D) (2/3 √2)

Tardigrade Admin · Accepted Answer

A=(A0/2t / T1 / 2) ⇒ (A/A0) =2-t / T1 / 2=2-(150/100) =2-3 / 2=(1/2 √2)

Q. The half-life of a radioactive nuclide is $100$ hours. The fraction of original activity that will remain after $150$ hours would be :

$1/2$

$\frac{1}{2 2}$

$\frac{2}{3}$

$\frac{2}{3 2}$

$A = \frac{A _{0}}{2 ^{t / T_{1/2}}} \Rightarrow \frac{A}{A _{0}}$
$= 2^{- t / T_{1/2}} = 2^{- \frac{150}{100}}$
$= 2^{- 3/2} = \frac{1}{2 2}$

Q. The half-life of a radioactive nuclide is 100 hours. The fraction of original activity that will remain after 150 hours would be :

1/2

22​1​

32​

32​2​