A radioactive isotope, A undergoes simultaneous decay to different nuclei as: \begin{array}{cc} \ce{A->P}&\,(t_{1/2}=9\ \mathrm h)\\ \ce{A->Q}&\,(t_{1/2}=4.5\ \mathrm h) \end{array}

Assuming that initially neither P nor Q was present, after how many hours, amount of Q will be just double to the amount of A remaining?

My solving:

I got the $t_{1/2}$ for $A$ as $T=\dfrac{T_1T_2}{T_1+T_2}\ =\pu{ 3 h}$

Now from Rutherford-soddy law the number of atoms left after $n$ half lives are,


and number of half lives $n=\dfrac{\text{Total time of Decay}}{\text{effective half life}}$

For isotope $A$, $\dfrac{N_A}{N_{0_A}}=\left(\dfrac{1}{2}\right)^{n}=\left(\dfrac{1}{2}\right)^{T/3}$ where $T$ is the common time of decay for $\mathrm A$ and $\mathrm Q$

For $\mathrm Q$ I had to assume that $\mathrm Q$ is also undergoing radioactive decay to form a relation between $\mathrm A$ and $\mathrm Q$

Now using the same formula as above $\dfrac{N_\mathrm Q}{N_{0_\mathrm Q}}=\left(\dfrac{1}{2}\right)^{T/4.5}=\left (\dfrac{1}{2}\right )^{2T/9}$

Now applying the condition in the question, $$\left (\frac{1}{2}\right)^{2T/9}=2\cdot \left (\dfrac{1}{2}\right)^{T/3} $$

Raising by power $9$ on both the sides,

$$\left (\dfrac{1}{2}\right )^{2T}=2^9\cdot \left (\frac{1}{2}\right )^{3T}$$

which implies,

$$ 2^{-2T}=2^{9-3T}$$


$$9-3T=-2T$$ $$ {\bbox[10px, border:2px solid red]{ T=9\ \mathrm h. }} $$

Although I have got an answer and it satisfies the condition I still think it's wrong because of the assumption made for $\mathrm Q$ and something else needed to be done to make the relation between $\mathrm A$ and $\mathrm Q$ Moreover, it's given in the question that initially neither $\mathrm P$ nor $\mathrm Q$ was present and I guess the assumption violates that.


2 Answers 2


The question has already been solved by Yashwini and the answer given is correct.$^2$ A more intuitive and specific to question explanation would follow here.

Now, the two reactions given are:

\begin{array}{cc} \require{cancel} \ce{A -> P} &(t_{1/2} = 9\,\mathrm h) \\ \ce{A -> Q} &(t_{1/2} = 4.5\,\mathrm h) \\ \end{array}

Now using the rate law, we get,

\begin{align} -\frac{\mathrm{d}[A]}{\mathrm{d}t}&=k_\mathrm P [A] \tag{1} \\ -\frac{\mathrm{d}[A]}{\mathrm{d}t}&=k_\mathrm Q [A] \tag{2} \\ \end{align}

The rate constant for a first order reaction having a half life of $t_{1/2}$ is defined as:

$$k=\frac{\ln 2}{t_{1/2}} \tag{3}$$

Now, substituting the given values of $t_{1/2}$ into the equations, we get $2k_\mathrm P = k_\mathrm Q$ (since $k\, \alpha \frac{1}{t_{1/2}})$

Now, intuitively since both reactions take place together, it would mean that for every one mole of P formed, two moles of Q forms. Therefore, for every mole of P formed, three moles of A react (since one mole is required for each mole of P and Q).

Now, we add the rate laws ($1$) and $(2)$, since the reactions take place simultaneously, to get:

$$-\frac{\mathrm{d}[A]}{\mathrm{d}t}=(k_\mathrm P +k_\mathrm Q) [A] \tag{4} $$

Now, since using the relation between $k_\mathrm{P}$ and $k_\mathrm{Q}$, we get $k_\mathrm{P} + k_\mathrm{Q} = 3k_\mathrm{P}$

Therefore using the integrated rate law for a first order reaction on equation $(4)$, we get:

$$A=A_0e^{-3k_\mathrm Pt} $$

Now, the amount of $A$ used here would be $A_0 -A$, and we get that value to be:

$$A_\text{used}=A_0\left(1-e^{-3k_\mathrm Pt}\right)$$

Now, as we have previously noted, for every three moles of A used, two moles Q are formed. This means that the amount of Q now in the mixture would be two thirds of $A_\text{used}$. Therefore the amount of Q would be:

$$Q=\frac{2A_0\left(1-e^{-3k_\mathrm Pt}\right)}{3}$$

Now, we are given the condition, $Q = 2A$, substituting values of $Q$ and $A$ into the given relation we get:

$$\begin{align} \frac{\cancel{2A_0}\left(1-e^{-3k_\mathrm Pt}\right)}{3} &= \cancel{2A_0}\left(e^{-3k_\mathrm Pt}\right) \\ \implies 1 -e^{-3k_\mathrm Pt} &= 3e^{-3k_\mathrm Pt} \\ \implies 4e^{-3k_\mathrm Pt} &= 1 \end{align}$$

Solving for $t$, we get:

\begin{align} 3k_\mathrm Pt&=2\ln 2 \\ \\ t&=\frac{2\ln 2}{3k_\mathrm P}\\ \end{align}

Now, using equation $(3)$, we get the rate constant $k_\mathrm P$ to be $\frac{\ln 2}{9}$. Substituting this value into the expression for time, we get:

$$t=\frac{2 \cancel{\ln 2}}{\cancel{3} \frac{\cancel{\ln 2}}{\cancelto{3}{9}}}$$

Therefore, time taken for this condition to happen is:

$$t=2\times 3 = 6\ \mathrm h$$

  • 3
    $\begingroup$ [2]: Re-formating said answer was an effort of about 90 minutes. $\endgroup$
    – Trish
    Aug 16, 2020 at 15:36

Parallel or side reactions of the first order: Concept

$$\require{cancel}\\ \ce{A ->[k_1] B} \ \ t=0\\ \ce{A ->[k_2] C} \ \ t=t$$ $$-\frac{\mathrm d[A]}{\mathrm dt}=k_1[A] + k_2[A] $$ $$-\frac{\mathrm d[A]}{\mathrm dt} = k_\text{eff} [A] \land k_\text{eff}=k_1+k_2$$

Effective order=1

$$\left(t_{1/2}\right)_\text{eff}=\frac {\ln 2}{k_\text{eff}} $$

$$\frac 1 {(t_{1/2})_\text{eff}}=\frac {1}{(t_{1/2})_{1}} + \frac {1} {(t_{1/2})_{2}} $$

$$A_\text{eff}\mathrm e^{-E_\mathrm a/(RT)}=(A_1+A_2)\mathrm e^{-E_\mathrm a/(RT)}$$

Differentiate with regards to $T$,

$${\frac{E_\mathrm a}{RT^2}}\cdot k_\text{eff}=\frac{(E_\mathrm a)_1 k_1}{RT^2}+\frac{(E_\mathrm a)_2 k_2}{RT^2}$$

$$(E_\mathrm a)_\text{eff}=\frac{(E_\mathrm a)_1 k_1 +(E_\mathrm a)_2 k_2}{k_\text{eff}}$$

$$[A]_\mathrm t=[A]_0\mathrm e^{-k_\text{eff}t}$$

$$a_t=a_0\mathrm e^{-(k_1+k_2)t}$$

$$\frac{\mathrm d[B]}{\mathrm dt}=k_1[A]=k_1a_0\mathrm e^{-(k_1+k_2)t}$$

$$\int\limits_{0}^{b_t}\mathrm d[B]=k_1 a_0 \int\limits_0^t\mathrm e^{-(k_1+k_2)t}\,\mathrm dt$$

$$b_t=\frac{k_1 a_0}{-(k_1+k_2)}[\mathrm e^{-(k_1+k_2)t}]_0^t$$

$$b_t=\frac{k_1 a_0}{k_1+k_2}(1-\mathrm e^{-(k_1+k_2)t}) $$


$$c_t=\frac{k_2 a_0}{k_1+k_2}(1-\mathrm e^{-(k_1+k_2)t})$$


  • proportion of $B=\frac{[B]}{x}=\frac {k_1}{k_1+k_2}$ [times 100 for percentage]
  • proportion of $C=\frac{[C]}{x}=\frac {k_2}{k_1+k_2}$ [times 100 for percentage]

The actual problem

\begin{align} &\ce{A->[\textit{k}_1]P} &k_1 &= \frac{\ln 2}{t_{1/2}} = \frac{\ln 2}{9} \ \text{hr}^{-1} \\ &\ce{A->[\textit{k}_2]Q} &k_2 &= \frac{\ln 2}{t_{1/2}} = \frac{2 \ln2}{9}\ \text{hr}^{-1}\\ \end{align}

$$Q_t=\frac{k_2a_0}{k_1+k_2}(1-\mathrm e^{-(k_1+k_2)t})=2A_t$$

$$\frac{k_2\cancel{a_0}}{k_1+k_2}\mathrm {(1-e^{-(k_1+k_2)t})}=2\cancel{a_0}\mathrm e^{-(k_1+k_2)t}$$

$$\frac{\cancel 2}{3}(1-\mathrm e^{-k_\text{eff}t})=\cancel 2\mathrm e^{-k_\text{eff}t}$$

$$\mathrm e^{-k_\text{eff}t} = \frac {1} {4}$$

$$\implies k_\text{eff}t = \ln 4 = \frac {3\ln 2}{9} t$$

$$\implies t= 6\mathrm h$$

So that gives the answer as 6 h.

  • 4
    $\begingroup$ For formatting, See here and here. For a more detailed MathJax guide, look here, minor other details $\endgroup$ Aug 16, 2020 at 10:58
  • $\begingroup$ Yashwini, though your solution is correct when applied to chemical reactions and the kinetics of radioactive decay, your reference to the temperature is problematic and may cause confusion and it is unnecesary.Even though you can model a nuclear reaction as a chemical reaction, the activation energy for these reactions are so high that temperature has no effect. I reccommend you to omit it from your explanation. $\endgroup$
    – PAEP
    Aug 22, 2020 at 18:48

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