
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
233.1k+ views
Hint: 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.
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.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding the Electric Field of a Uniformly Charged Ring

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions For Class 12 Chemistry Chapter 1 Solutions (2025-26)

Solutions Class 12 Chemistry Chapter 1 CBSE Notes - 2025-26

NCERT Solutions For Class 12 Chemistry Chapter 4 The d and f Block Elements (2025-26)

Biomolecules Class 12 Chemistry Chapter 10 CBSE Notes - 2025-26

NCERT Solutions For Class 12 Chemistry Chapter 10 Biomolecules (2025-26)

