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

The correct option is 1/(2√2).

To find the fraction of the original radioactive activity remaining after a certain period of time, we can use the law of radioactive decay.

The fraction of the active nuclei (or the fraction of the activity) remaining after time
$t$
is given by the formula:

$\frac{A}{A_0}={\left(\frac{1}{2}\right)}^n$

where:
•
$A$
is the activity at time
$t$
.
•
$A_0$
is the initial activity.
•
$n$
is the number of half-lives that have elapsed.

The number of half-lives
$n$
is calculated as:

$n=\frac{t}{T_{1/2}}$

Given in the problem:
• Half-life (
$T_{1/2}$
) = 100 hours
• Time elapsed (
$t$
) = 150 hours

First, let us calculate the number of half-lives,
$n$
:

$n=\frac{150}{100}=\frac{3}{2}$

Now, substitute this value into the fraction formula:

$\frac{A}{A_0}={\left(\frac{1}{2}\right)}^{3/2}$

We can simplify this expression step-by-step:

${\left(\frac{1}{2}\right)}^{3/2}=\frac{1^{3/2}}{2^{3/2}}=\frac{1}{\sqrt{2^3}}$

Since
$2^3=8$
, we have:

$\sqrt{8}=\sqrt{4\times 2}=2\sqrt{2}$

Thus, the fraction of original activity that remains is:

$\frac{A}{A_0}=\frac{1}{2\sqrt{2}}$