Question

half life of a radioactive material is K the fraction that would remain after k/2 is

Answer 1

Answer:

0.7

Explanation:

The radioactive decay is first order change.

The relation between t1/2 and rate constant (k) for first order reaction is

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

We have been given that t1/2 is K

$K= \frac{0.693}{k}$

so rate constant 'k' will be

$k= \frac{0.693}{K}$

Now we know that the integrated equation of first order is

$A=A_{0}e^{-kt}$

where A is the fraction of reactant is remained and A0 is the initial concentration of reactant.

putting the value of rate constant

$k= \frac{0.693}{K}$

and t=K/2

$A=A_{0}e^{-\frac{0.693}{K}{\frac{K}{2}}$

=0.7

Answer 2

Answer: $\frac{1}{1.414}$

Explanation: Now, to calculate the amount left , we use the formula:

$a=\frac{a_o}{2^n}$

where,

a = amount of reactant left after n-half lives = ?

$a_o$ = Initial amount of the reactant = $a_0$

n = number of half lives  =$\frac{\text {Given time}}{\text [half life}}=\frac{k/2}{k}= \frac{1}{2}$

Putting values in above equation, we get:

$a=\frac{a_0}{2^\frac{1}{2}$

$a=\frac{a_0]{1.414}$

Thus the amount of reactant left after $\frac{k}{2}$ minutes is $\frac{1}{4.414}$ of original concentration.