Question

There is a 30 g sample of be-11 it has a half life of about 14 seconds how much will be left after 28 seconds

Answer 1

the answer is 3.75 g

^_~

Answer 2

Answer : The amount left after 28 seconds will be, 7.505 grams

Solution :

As we know that the radioactive decays follow the first order kinetics.

First we have to calculate the half life of a Be-11.

Formula used :
`t_(1/2)=(0.693)/(k)`

Putting value of 'half-life' in this formula, we get the value of 'k'.

`14s=(0.693)/(k)`

`k=0.0495s^(-1)`

The expression for rate law for first order kinetics is given by :

`k=(2.303)/(t)\log(a)/(a-x)`

where,

k = rate constant = 0.0495

t = time taken for decay process = 28 s

a = initial amount of the reactant = 30 g

a - x = amount left after decay process = ?

Now put all the given values in this formula, we get the amount left after decay.

`0.0495=(2.303)/(28)\log(30)/(a-x)`

`a-x=7.505g`

Therefore, the amount left after 28 seconds will be, 7.505 grams.