Question

The half life period of a first order radioactivity decay of an element is 5 days. The time (in days) taken for 8g of this element to decay to 1g is:
a) 15
b) 25
c) 40
d) 10

Answer 1

Hint: From half life we can understand that the question is talking about half of the total decay is completed. And by applying the formula of half life for 1st order reaction we can solve the problem.

Step by step solution: Radioactive decay (also known as nuclear decay, radioactivity, radioactive disintegration or nuclear disintegration) is the process by which an unstable atomic nucleus loses energy by radiation. In other terms radioactive decay is the spontaneous breakdown of an atomic nucleus resulting in the release of energy and matter from the nucleus. For this decay half life is the time taken for the radioactivity of a specified isotope to fall to half its original value. From the derivation of 1st order reaction we know that formula of half life is:
\[{{t}_{\tfrac{1}{2}}}=\dfrac{0.693}{k}\]
Where, \[{{t}_{\tfrac{1}{2}}}\]is the half life,
\[k\] is the rate constant.
By using the half life data we will find the value of rate constant.
\[k=\dfrac{0.693}{{{t}_{\tfrac{1}{2}}}}\]
\[k=\dfrac{0.693}{5}\] \[da{{y}^{-1}}\]
     = 0.1386 \[da{{y}^{-1}}\]
Now we know that the formula for the time of 1st order reaction :
\[t=\dfrac{2.303}{k}\times \log (\dfrac{{{C}_{1}}}{{{C}_{2}}})\]
Here, \[{{C}_{1}}\]is initial concentration and \[{{C}_{2}}\]is final concentration.
In the question: \[{{C}_{2}}\]=1 and \[{{C}_{1}}\]=8
By putting this data and value of k in time equation:
\[t=\dfrac{2.303}{0.1386}\times \log (\dfrac{8}{1})\]
 t = 15 days

So, the correct answer is “A”.

Note: Unit of time constant for 1st order reaction is \[tim{{e}^{-1}}\]. And units of time can be anything like sec, min or days etc. Radioactive decay occurs in unstable atomic nuclei.