Question

the half life period of radioisotope is 20 s.starting with 0.2 g of the isotope ,the amount left after 2mins

Answer 1

Answer: 0.0055 grams

Explanation:

Radioactive isotopes follows first order kinetics.

Half-life of isotope = 20 sec

First we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{t_{\frac{1}{2}}}$

$k=\frac{0.693}{20}=0.03{s}^{-1}$

Now we have to calculate the amount left after 2 mins

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant  = $0.03{s}^{-1}$

t = time for decomposition  = $2min=2\times 60=120sec$

a = initial amount of the reactant  = 0.2 g

a - x = amount left after decay process  = ?

Now put all the given values in above equation, we get

$120=\frac{2.303}{0.03}\log\frac{0.2}{a-x}$

$(a-x)=0.0055g$

Thus the amount left after 2 minutes is 0.0055 grams.

Answer 2

Hello!

The half - life period of radioisotope is 20 seconds. Starting with 0.2 g of the isotope, the amount left after 2 minutes is ?

• We have the following data:

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

P (Half-life) = 20 sec

mo (initial mass) = 0.2 g

T (Elapsed time for sample reduction) = 2 minutes = 120 sec

x (number of periods elapsed) = ?

m (final mass after time T) = ? (in grams)

• Let's find the number of periods elapsed, in formula the half-life (P) of the radioactive sample, let's see:

$T = x*P$

$120 = x*20$

$20\:x =120$

$x = \dfrac{120}{20}$

$\boxed{x = 6}$

• Now, Let's find the final mass after time T, let us see:

$m = \dfrac{m_o}{2^x}$

$2^x = \dfrac{m_o}{m}$

$2^6 = \dfrac{0.2}{m}$

$64 = \dfrac{0.2}{m}$

$64\:m = 0.2$

$m = \dfrac{0.2}{64}$

$m = 3.125*10^{-3}$

$\boxed{\boxed{m = 0.003125\:g}}\:\:\:\:\:\:\bf\green{\checkmark}$

Answer:

0.003125 g

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