Question

The radioactive isotope 32P decays by first-order kinetics and has a half-life of 14.3 days. How long does it take for 75.0% of a sample of 32P to decay?

Answer 1

Answer : The time passed in days is 28.6 days.

Explanation :

Half-life = 14.3 days

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

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

$k=\frac{0.693}{14.3\text{ days}}$

$k=4.85\times 10^{-2}\text{ days}^{-1}$

Now we have to calculate the time passed.

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  = $4.85\times 10^{-2}\text{ days}^{-1}$

t = time passed by the sample  = ?

a = let initial amount of the reactant  = 100 g

a - x = amount left after decay process = 100 - 75.0 = 25.0 g

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

$t=\frac{2.303}{4.85\times 10^{-2}}\log\frac{100}{25.0}$

$t=28.6\text{ days}$

Therefore, the time passed in days is 28.6 days.

Answer 2

The time it takes for 75% of the sample to decay is required.

Time taken for the required decay is 28.6 days.

$t_{1/2}$ = Half-life of sample = 14.3 days

$N$ = Amount of sample left to decay = $100-75=25\%$

$N_0$ = Decay at time $t=0$ = 100%

Decay constant

$\lambda=\dfrac{\ln 2}{t_{1/2}}$

Radioactive decay is given by

$N=N_0e^{-\dfrac{\ln 2}{t_{1/2}}t}\\\Rightarrow \ln\dfrac{N}{N_0}=-\dfrac{\ln 2}{t_{1/2}}t\\\Rightarrow t=\dfrac{\ln\dfrac{N}{N_0}}{-\dfrac{\ln 2}{t_{1/2}}}\\\Rightarrow t=\dfrac{\ln\dfrac{25}{100}}{-\dfrac{\ln 2}{14.3}}\\\Rightarrow t=28.6\ \text{days}$

Time taken for the required decay is 28.6 days.

Learn more:

https://brainly.com/question/3949343?referrer=searchResults