Question

A radioactive substance contains 10000 nuclei and its half life period is 20 days. the no of nuclei present at the end of 10 days are

Answer 1

The remaining amount after a radioactive decay is given by-

$N(t) = N_{0}e^{\frac{-0.693t}{t_{\frac12}}}$

where
N(t) is the amount remaining after time t
$N_0$ is the initial amount
$t_{\frac12}$ is the half life

Putting the values-
$N(t) = 10000e^{(\frac{-0.693 \times 10}{20})}$
or, $N(t) = 10000 \times 0.707$
or, $N(t) = 7070$ (appr.)

So, there will be 7070 atoms left (approximately)

Answer 2

this is the ans is 7070 atom left