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 : [tex]t_{1/2}=\frac{0.693}{k}[/tex]
Putting value of 'half-life' in this formula, we get the value of 'k'.
[tex]14s=\frac{0.693}{k}[/tex]
[tex]k=0.0495s^{-1}[/tex]
The expression for rate law for first order kinetics is given by :
[tex]k=\frac{2.303}{t}\log\frac{a}{a-x}[/tex]
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.
[tex]0.0495=\frac{2.303}{28}\log\frac{30}{a-x}[/tex]
[tex]a-x=7.505g[/tex]
Therefore, the amount left after 28 seconds will be, 7.505 grams.
