Ncert Exercises & Solutions

Sometimes a radioactive nucleus decays into a nucleus which itself is radioactive. An example is :

$^{38} Sulphur \to_{= 2.48 h}^{half - life}^{38} Cl \to_{= 0.62 h}^{half - life}^{38} Ar (stable)$

Assume that we start with 1000 $^{38} S$ nuclei at time t = 0. The number of $^{38} Cl$ is of count zero at t = 0 and will again be zero at t = $\infty$ . At what value of t, would the number of counts be a maximum?

Hint: The rate of disintegration depends on the active nuclei.

Step 1: Find the decay rate of two nuclei.

Consider the chain of two decays.

^{38} S \underset{2.48 h}{\to}^{38} Cl \underset{0.62 h}{\to}^{38} Ar

At time t, let

^{38} S

has

N_{1} (t)

active nuclei and

^{38} Cl

has

N_{2} (t)

active nuclei.

\frac{{dN}_{1}}{dt} = - λ_{1} N_{1} = rate of formation of {Cl}^{38} .

Also, \frac{{dN}_{2}}{dt} = - λ_{2} N_{2} + λ_{1} N_{1}

But, N_{1} = N_{0} e^{- λ_{1} t}

\frac{{dN}_{2}}{dt} = λ_{1} N_{0} e^{- λ_{1} t} - λ_{2} N_{2} . . . . . (i)

Step 2: Find the no. of counts of

^{38} Cl

Multiplying by

e^{λ_{2} t} dt

and rearranging

e^{λ_{2} t} {dN}_{2} + λ_{2} N_{2} e^{λ_{2} t} dt = λ_{1} N_{0} e^{(λ_{2} - λ_{1}) t} dt

Integrating both sides:

N_{2} e^{λ_{2} t} = \frac{N_{0} λ_{1}}{λ_{2} - λ_{1}} e^{(λ_{2} - λ_{1}) t} + C

Since at t = 0, N_{2} = 0, C = - \frac{N_{0} λ_{1}}{λ_{2} - λ_{1}}

So, N_{2} e^{λ_{2} t} = \frac{N_{0} λ_{1}}{λ_{2} - λ_{1}} (e^{(λ_{2} - λ_{1}) t} - 1) . . . . . . . . . . . . . . (ii)

Step 3: Find the condition when no. of counts of

^{38} Cl

is maximum.

For maximum count,

\frac{{dN}_{2}}{dt} = 0

λ_{1} N_{0} e^{- λ_{1} t} - λ_{2} N_{2} = 0 [From Eq . (i)]

\Rightarrow \frac{N_{0}}{N_{2}} = \frac{λ_{2}}{λ_{1}} e^{λ_{1} t} [From Eq . (ii)]

e^{λ_{2} t} - \frac{λ_{2}}{λ_{1}} \cdot \frac{λ_{1}}{(λ_{1} - λ_{1})} e^{λ_{1} t} [e^{(λ_{2} - λ_{1}) t} - 1] = 0

or e^{λ_{2} t} - \frac{λ_{2}}{(λ_{2} - λ_{1})} e^{λ_{2} t} + \frac{λ_{2}}{(λ_{2} - λ_{1})} e^{λ_{1} t} = 0

1 - \frac{λ_{2}}{(λ_{2} - λ_{1})} + \frac{λ_{2}}{(λ_{2} - λ_{1})} e^{(λ_{1} - λ_{2}) t} = 0

\frac{λ_{2}}{(λ_{2} - λ_{1})} e^{(λ_{1} - λ_{2}) t} = \frac{λ_{2}}{(λ_{2} - λ_{1})} - 1

e^{(λ_{1} - λ_{2}) t} = \frac{λ_{1}}{λ_{2}}

Step 4: Find the time when no. of counts of

^{38} Cl

is maximum.

t = \frac{(\log_{e} \frac{λ_{1}}{λ_{2}})}{(λ_{1} - λ_{2})}

= \frac{\log_{e} (\frac{2.48}{0.62})}{2.48 - 0.62}

= \frac{\log_{e} 4}{1.86} = \frac{2.303 \times 2 \times 0.3010}{1.86} (∵ λ = \frac{0.693}{T_{1 / 2}})

= 0.745 S

Note Do not apply directly the formula of radioactive. Apply formulae related to chain decay.

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