Question

i) what is meant by half life of a radioactive element?

ii) the half life of a radioactive substance is 30s. calculate the decay constant and time taken for the sample to decay by 3/4 of the initial value.​

Answer 1

Part (i)

Half life of a radioactive element:

Half life of a radioactive element is defined as the time during which half the number of atoms present initially in the sample of the element decay or it is the same during which the number of atoms left undecayed in the sample is half the total number of atoms present in the sample.

• It is represented by $\displaystyle\sf T_{\frac{1}{2}}$

From the equation $\displaystyle\sf N = N_0 \:e^{-\lambda t}$

At half life, $\displaystyle\sf t = T_{\frac{1}{2}}\;\:\;\text{and}\;\;\;N = \dfrac{N_0}{2}$

$\displaystyle\sf \therefore \frac{N_0}{2} = N_0 \: e^{-\lambda T_{\frac{1}{2}}}$

$\displaystyle\sf \implies \dfrac{1}{2} = e^{-\lambda T_{\frac{1}{2}}}$

$\displaystyle\sf \implies e^{\lambda T_{\frac{1}{2}}} = 2$

• on taking log both sides

$\displaystyle\sf \lambda T_{\frac{1}{2}} = \log_e 2$

$\displaystyle\sf T_{\frac{1}{2}} = \dfrac{\log_e 2}{\lambda}$

$\displaystyle\sf = \dfrac{\log_10 2\times 2.303}{\lambda}$

$\displaystyle\sf = \dfrac{0.3010\times 2.303}{\lambda}$

After n half times, the number of atoms left undecayed is given by

$\displaystyle\sf N = N_0 \left(\frac{1}{2}\right)^n$

$\displaystyle\sf T_{\frac{1}{2}} = \dfrac{0.6932}{\lambda}$

Average life of a radioactive element:

Average life of a radioactive element can be obtained by calculating the total life time of all atoms of the element and dividing it by total number of the atoms present initially in the sample of the element.

Average life or mean life of radioactive element is

$\displaystyle\sf \tau = \dfrac{\text{total life time of all atoms}}{\text{total number of atoms}}$

$\displaystyle\sf \tau = \int\limits_0^{N_0} \dfrac{tdN}{N_0}$

Now, when N = N_0, t = 0, and when N = 0, t = ∞.

$\displaystyle\sf \tau = \int\limits_{\infty}^{0} \dfrac{-\lambda N_0 e^{-\lambda t} dt\times t}{N_0}\;\;\;\;\bf \boldsymbol[ \boldsymbol\because dN = -\boldsymbol\lambda\boldsymbol(N_0e^{-\boldsymbol\lambda t}\boldsymbol) dt\boldsymbol]$

$\displaystyle\sf = \lambda \int\limits_0^{\infty} te^{-\lambda t} dt$

$\displaystyle\sf = \lambda \left[ \left\{ t \dfrac{e^{-\lambda t}}{-\lambda}\right\}_0^{\infty} - \int\limits_0^{\infty} \dfrac{e^{-\lambda}}{\lambda} dt\right]$

$\displaystyle\sf = \lambda \left( 0+ \dfrac{1}{\lambda}\int\limits_0^{\infty} e^{-\lambda t} dt\right)$

$\displaystyle\sf = \int\limits_0^{\infty} e^{-\lambda t} dt$

$\displaystyle\sf = \left[ \dfrac{e^{-\lambda t}}{-\lambda} \right]^{\infty}_0$

$\displaystyle\sf = 0-\dfrac{1}{-\lambda}$

$\displaystyle\sf = \dfrac{1}{\lambda}$

$\displaystyle\sf \tau = \dfrac{1}{\lambda} = \dfrac{1}{0.6931/T_{\frac{1}{2}}}$

$\displaystyle\sf = 1.44 T_{\frac{1}{2}}$

Part (ii)

Given, $\displaystyle\sf \lambda = 0.3465 \:(day)^{-1}$

According to radioactive decay law, we have

$\displaystyle\sf R = R_0 e^{-\lambda t}$

$\displaystyle\sf \implies \dfrac{R_0 \times 75}{100} = R_0 \; e^{-0.3465t}$

$\displaystyle\sf \implies \dfrac{3}{4} = e^{-0.3465t}$

$\displaystyle\sf \implies t=0.830\;s$

Also given that, $\displaystyle\sf T_{\frac{1}{2}} = 30\;s$

a) $\displaystyle\sf \lambda = ?$

$\displaystyle\sf \because T_\frac{1}{2} = \dfrac{0.693}{\lambda}$

$\displaystyle\sf \implies \lambda = \dfrac{0.693}{T_{\frac{1}{2}}}$

$\displaystyle\sf = \dfrac{0.693}{30}$

$\displaystyle\sf\boxed{\sf \therefore \lambda = 0.0231\:s^{-1}}$

b) $\displaystyle\sf \because N = N_0\left(\dfrac{1}{2}\right)^n$

• where n = number of half lives
• N = no. of undisintegrated nuclei present in the sample
• N_0 = original number of undisintegrated atoms

Here, $\displaystyle\sf N = N_0 - \dfrac{3}{4} N_0$

$\displaystyle\sf N = \dfrac{1}{4} N_0$

$\displaystyle\sf \implies N = N_0 \left(\frac{1}{2}\right)^n$

$\displaystyle\sf \dfrac{N_0}{4} = N_0\left(\dfrac{1}{2}\right)^n$

$\displaystyle\sf \implies \left(\dfrac{1}{2}\right)^2 = \left(\frac{1}{2}\right)^n$

$\displaystyle\sf \implies n = 2$

But number of life times

$\displaystyle\sf 2 = \dfrac{Total\;time\:taken}{30}$

$\displaystyle\boxed{\sf Total\:time = 60\:s = 1\:minute}$