Question

80% of the radioactive nuclei present in a sample are found to remain undecayed after one day. The percentage of undecayed nuclei left after two days will be ______________.
A. 64
B. 20
C. 46
D. 80

Answer 1

Radioactive decay is the spontaneous breakdown of an atomic nucleus resulting in the release of energy and matter from the nucleus. Radioactivity follows first order i.e. the Rate of decay of the nuclei at a given time is directly proportional to concentration of the nuclei at that point of time.

Complete Step by Step Solution: By using the formula written below, we can solve this question:
$K=\dfrac{2.303}{t}\log \left( \dfrac{{{N}_{o}}}{N} \right)$ ……. (I)
Where, ${{N}_{o}}$ is the initial amount of substance before the decay process starts.
N is the final amount of substance left after the decay process ends i.e. undecayed sample.
t is the time period in which the decay process continues.
According to the question, 80% of the radioactive nuclei present in a sample are found to remain undecayed after one day.
Hence put all values in equation (I)
$K=\dfrac{2.303}{1}\log \left( \dfrac{100}{80} \right)$ ………(II)
Now, the percentage of undecayed nuclei left after two days will be, N1
$K=\dfrac{2.303}{t}\log \left( \dfrac{100}{{{N}_{o}}} \right)$ ………. (III)
Compare equation (II) and equation (III):
$\dfrac{2.303}{1}\log \left( \dfrac{100}{80} \right)=\dfrac{2.303}{2}\log \left( \dfrac{100}{{{N}_{1}}} \right)$
${{\left( \dfrac{5}{4} \right)}^{2}}=\dfrac{100}{{{N}_{1}}}$
${{N}_{1}}$ = 64.
Hence, the percentage of undecayed nuclei left after two days will be 64.

Hence, option A is the correct answer.

Note:
Always remember, the formula $K=\dfrac{2.303}{t}\log \left( \dfrac{{{N}_{o}}}{N} \right)$ , where ${{N}_{o}}$ is initial amount of substance, N is the final amount of substance left, t is the time period and K is the constant.
Remember that a radioisotope has unstable nuclei that does not have enough binding energy to hold the nucleus together.
Radioisotopes would like to be stable isotopes so they are constantly changing to try and stabilize.