Question

If a 900.0-g sample of radium-226 decays to 225.0 g of radium-226 remaining in 3,200 years, what is the half-life of radium-226?

Answer 1

Answer : The half-life of radium-226 is, 1600.46 years

Solution : Given,

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

So, the expression for rate law for first order kinetics is given by :

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

where,

k = rate constant  = ?

t = time taken for decay process  = 3200 years

a = initial amount of the radium-226 = 900 g

a - x = amount left after decay process  = 225 g

Putting values in above equation, we get the value of rate constant.

$k=\frac{2.303}{3200}\log\frac{900}{225}=4.33\times 10^{-3}Year^{-1}$

Now we have to calculate the half life of a radium-226.

Formula used :

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

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

$t_{1/2}=\frac{0.693}{4.33\times 10^{-3}Year^{-1}}=1600.46\text {Years}$

Therefore, the half-life of radium-226 is, 1600.46 years

Answer 2

After 3200 years, the amount remaining of the original 900 g is 225/900 = 1/4.
Each half-life is enough time to reduce the amount of a radioactive material by a factor of 2.
1/4 is a reduction by two factors of 2. (1/2 * 1/2 = 1/4)
So 3200 years is two half-lives, so one half-life is 1600 years. Source(s):