Question 195: Atoms & Nuclei

Question

A radioactive material decays by simultaneous emissions of two particles with half lives of 1400 years and 700 years respectively. What will be the time after which one third of the material remains ? (Take ln 3 = 1.1)

Accepted Answer

The given situation can be shown as
Here, radioactive material X is decayed into two particles Y and Z with their respective decay constant, $$\lambda$$
a
and $$\lambda$$
b
. It means that
$$\because$$ $$\lambda  = {{\ln 2} \over {{t_{1/2}}}}$$
where, t
1/2
(a) = 700 yr
and t
1/2
(b) = 1400 yr
$${\lambda _a} = {{\ln 2} \over {700}}y{r^{ - 1}}$$ and $${\lambda _b} = {{\ln 2} \over {1400}}y{r^{ - 1}}$$
$$\because$$ $${\lambda _{total}} = {\lambda _a} + {\lambda _b}$$
$$ = \left( {{{\ln 2} \over {700}} + {{\ln 2} \over {1400}}} ight)y{r^{ - 1}} = \ln 2\left( {{1 \over {700}} + {1 \over {1400}}} ight)y{r^{ - 1}}$$
$$ = \left( {{{3\ln 2} \over {1400}}} ight)y{r^{ - 1}}$$
Suppose the initial number of radioactive nuclei was N
0
.
$$	herefore$$ $$N = {N_0}{e^{ - \lambda t}}$$
where, N = number of nuclei present at time = t and N
0
= number of nuclei present at time = 0
$$ \Rightarrow {{{N_0}} \over 3} = {N_0}{e^{ - \lambda t}}$$ or $${{{N_0}} \over 3} = {N_0}{N_0}{e^{ - {\lambda _{total}}t}} \Rightarrow {1 \over 3} = {e^{ - {\lambda _{total}}t}}$$
Taking log on both the sides of above equation, we get
$$\ln \left( {{1 \over 3}} ight) = \ln ({e^{ - {\lambda _{total}}t}})$$
$$ \Rightarrow \ln \left( {{1 \over 3}} ight) =  - {\lambda _{total}}t$$
$$ \Rightarrow 1.1 = {{3 	imes 0.693} \over {1400}} 	imes t$$
$$\Rightarrow$$ t $$\approx$$ 740 yr

Answer

740 years

Answer

1110 years

Answer

700 years

Answer

340 years

Atoms & Nuclei

A radioactive material decays by simultaneous emissions of two particles with half lives of 1400 years and 700 years respectively. What will be the time after which one third of the material remains ? (Take ln 3 = 1.1)

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