Question

There are 4√2×10^6 radioactive nuclei in a Given radioactive sample.if the half life of the sample is 20 s , how many nuclei will decay in 10 s?

Answer 1

Since radioactive decay is first order reaction.

K= ln2/20

Then use k in

R(remaining)= R(initial) e^(-k10)

= 3.999*10^6

Substrate from initial value

=1.657*10^6

Answer 2

Given:

• nuclei number $N_{0}=4\sqrt{2}\times 10^{6}$
• half life, T = 20 s

To find: number of nuclei decaying in 10 s = ?

Solution:

We know that, decay constant,

$\quad\quad\lambda = \frac{0.693}{T}$

$\quad\quad\quad =\frac{0.693}{20}$

$\quad\quad\quad =0.03465$

∴ the number of nuclei left after 10 s is given by

$\quad N=N_{0}e^{-\lambda t}$

$\to N=4\sqrt{2}\times 10^{6}\times e^{-0.03465\times 10}$

$\to N=4\times 10^{6}$

∴ the number of nuclei decaying in 10 s is

$\quad N_{0}-N$

$=(4\sqrt{2}\times 10^{6})-(4\times 10^{6})$

$=1.66\times 10^{6}$