Question

A radioactive isotope decomposes according to the first order with a half-life period of 15 hrs. 80% of the sample will decompose in –
(A) $15 \times 0.8{\text{ hr}}$
(B) $15 \times (\log 8){\text{ hr}}$
(C) $15 \times (\dfrac{{\log 5}}{{\log 2}}){\text{ hr (or 34}}{\text{.83) hr}}$
(D) $15 \times \dfrac{{10}}{8}{\text{ hr}}$

Answer 1

An isotope of an element, is the same entity but having a different mass number, sometimes the isotopes are unstable so these are called radioactive isotopes or radioisotopes. Since the isotopes are unstable, they dissociate or decay, the time at which the decay is around 50% of its original value this time is known as half-life time.

Formula used:(1) ${{\text{t}}_{\dfrac{1}{2}}}{\text{ = }}\dfrac{{{\text{0}}{\text{.693}}}}{{{\lambda }}}$ , where ${{\text{t}}_{\dfrac{1}{2}}}$ = half-life time (given in the question) and ${{\lambda }}$ = decaying constantly
(2) $-$ ${\text{t = }}$ $\dfrac{{\text{1}}}{{{\lambda }}} \times 2.303\log \dfrac{{{{\text{A}}_{\text{f}}}}}{{{{\text{A}}_{\text{i}}}}}$ where ${{\text{A}}_{\text{f}}}$ is the final amount of substance left, after the decay
${{\text{A}}_{\text{i}}}$is the initial amount of substance taken

Complete step-by-step solution: According to the question 80% of the radioactive decay has occurred, that means if we consider initially 100% of a substance was taken then, we are left with 20% of that substance (since 100 – 80 = 20)
Therefore, the value of ${{\text{A}}_{\text{f}}}$ = 20 and ${{\text{A}}_{\text{i}}}$ = 100,
But to calculate the time taken, we need to determine the value of decaying constant (${{\lambda }}$), and the formula for the decaying constant can be written as;
$\Rightarrow$ ${{\text{t}}_{\dfrac{1}{2}}}{\text{ = }}\dfrac{{{\text{0}}{\text{.693}}}}{{{\lambda }}}$
$\Rightarrow$ ${{\lambda = }}\dfrac{{{\text{0}}{\text{.693}}}}{{{{\text{t}}_{\dfrac{1}{2}}}}}$ (substituting the value of t ½ will give us the value of decaying constant)
$\Rightarrow$ ${{\lambda = }}\dfrac{{{\text{0}}{\text{.693}}}}{{15}}$
$\Rightarrow$ ${{\lambda = 0}}{\text{.0462}}$
Now, if we substitute all the values in the mentioned formula (ii) of Time(t) we get,
$\Rightarrow$ $-$ ${\text{t = }}$ $\dfrac{{\text{1}}}{{{\lambda }}} \times 2.303\log \dfrac{{{{\text{A}}_{\text{f}}}}}{{{{\text{A}}_{\text{i}}}}}$
$\Rightarrow$ $-$ ${\text{t = }}$ $\dfrac{{\text{1}}}{{0.0462}} \times 2.303\log \dfrac{{20}}{{100}}$
$\Rightarrow$ $-$ ${\text{t = }}$ $\dfrac{{\text{1}}}{{0.0462}} \times 2.303\log \dfrac{2}{{10}}$
$\Rightarrow$ $-$ ${\text{t = }}$ $\dfrac{{\text{1}}}{{0.0462}} \times 2.303(\log 2 - \log 10)$ [here we have used the logarithmic expression, ${\text{log}}\dfrac{{\text{a}}}{{\text{b}}} = \log {\text{a }} - \log {\text{b}}$]
$\Rightarrow$ $-$ ${\text{t = }}$ $\dfrac{{\text{1}}}{{0.0462}} \times 2.303(0.30102 - 1)$ [the value of log 2 = 0.3010 and the value of –log 10 = $-$1]
$\Rightarrow$ $-$ ${\text{t = }}$ $\dfrac{{\text{1}}}{{0.0462}} \times 2.303( - 0.698)$ ( now carry out the simple arithmetic operation, and get to the required solution)
$\Rightarrow$ $-$ ${\text{t = }}$ $\dfrac{{\text{1}}}{{0.0462}} \times 2.303( - 0.698)$
$\Rightarrow$ $-$ ${\text{t = }}$ $- \dfrac{{\text{1}}}{{0.0462}} \times 1.607494$
$\Rightarrow$ $-$ ${\text{t = }}$ $- 34.78$
$\Rightarrow$ ${\text{t = }}$ $34.78$ (the two negative signs will cancel each other)
$\Rightarrow$ ${\text{t = }}$ $34.8$ hr
The obtained value of time (t) is very much close to the option (C)

Hence, the correct answer is an option (C) i.e. $15 \times (\dfrac{{\log 5}}{{\log 2}}){\text{ hr (or 34}}{\text{.83) hr}}$.

Note: The correct answer has the logarithmic part (i.e., $\dfrac{{\log 5}}{{\log 2}}$), which is not similar to the logarithmic value (i.e., $\log \dfrac{2}{{10}}$)we have to get during our solution, however, the solution is correct, this is because the same solution can be done by using another variant of the same formula;
$\Rightarrow$ ${\text{t = }}$ $\dfrac{{\text{1}}}{{t{\lambda }}} \times 2.303\log \dfrac{{\text{a}}}{{{\text{a}} - {\text{x}}}}$, notice the change in the negative sign has changed the equation.