
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
153.9k+ 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 2022 (June 29th Shift 2) Maths Question Paper with Answer Key

JEE Main 2023 (January 25th Shift 1) Maths Question Paper with Answer Key

JEE Main 2022 (July 29th Shift 1) Maths Question Paper with Answer Key

JEE Main 2022 (July 26th Shift 2) Chemistry Question Paper with Answer Key

JEE Main 2022 (June 26th Shift 2) Maths Question Paper with Answer Key

JEE Main 2022 (June 29th Shift 1) Physics Question Paper with Answer Key

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Degree of Dissociation and Its Formula With Solved Example for JEE

Displacement-Time Graph and Velocity-Time Graph for JEE

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Other Pages
NCERT Solutions for Class 12 Chemistry Chapter 1 Solutions

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

NCERT Solutions for Class 12 Chemistry Chapter 2 Electrochemistry

NCERT Solutions for Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes

Solutions Class 12 Notes: CBSE Chemistry Chapter 1

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More
