JEE Exam  >  JEE Questions  >  Rn decays into Po by emitting an alpha-partic... Start Learning for Free
Rn decays into Po by emitting an alpha-particle with half life of 4 days. A sample contains 6.4 × 10^(10) atoms of Rn. After 12 days, the number of atoms of Rn left in the sample will be a) 3.2 × 10^(10) b) 0.53 ×10^(10) c) 2.1 × 10^(10) d) 0.8 × 10^(10)?
Verified Answer
Rn decays into Po by emitting an alpha-particle with half life of 4 da...
Correct option is (d).
This question is part of UPSC exam. View all JEE courses
Most Upvoted Answer
Rn decays into Po by emitting an alpha-particle with half life of 4 da...
Use N=N′×e^[-(Lambda) t]
=0.8×10^10

or as radioactive decay is first order reaction, after 3 half lives , remaining amount of sample will be 1/8th of the initial. hence 0.8 ×10 ^-10
Community Answer
Rn decays into Po by emitting an alpha-particle with half life of 4 da...
Understanding the Problem:
- Rn decays into Po by emitting an alpha-particle with a half-life of 4 days.
- A sample contains 6.4 × 10^(10) atoms of Rn.

Calculating Decay:
- Since the half-life of Rn is 4 days, after 4 days, half of the Rn atoms will decay.
- After 4 days, the number of Rn atoms left = 6.4 × 10^(10) / 2 = 3.2 × 10^(10).
- After another 4 days (total 8 days), half of the remaining Rn atoms will decay.
- Rn atoms left after 8 days = 3.2 × 10^(10) / 2 = 1.6 × 10^(10).

Finding Atoms After 12 Days:
- After 12 days, another half of the remaining Rn atoms will decay.
- Rn atoms left after 12 days = 1.6 × 10^(10) / 2 = 0.8 × 10^(10) = 8.0 × 10^(9).

Conclusion:
After 12 days, the number of atoms of Rn left in the sample will be 0.8 × 10^(10) or 8.0 × 10^(9). Therefore, the correct answer is option d) 0.8 × 10^(10).
Explore Courses for JEE exam
Rn decays into Po by emitting an alpha-particle with half life of 4 days. A sample contains 6.4 × 10^(10) atoms of Rn. After 12 days, the number of atoms of Rn left in the sample will be a) 3.2 × 10^(10) b) 0.53 ×10^(10) c) 2.1 × 10^(10) d) 0.8 × 10^(10)?
Question Description
Rn decays into Po by emitting an alpha-particle with half life of 4 days. A sample contains 6.4 × 10^(10) atoms of Rn. After 12 days, the number of atoms of Rn left in the sample will be a) 3.2 × 10^(10) b) 0.53 ×10^(10) c) 2.1 × 10^(10) d) 0.8 × 10^(10)? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Rn decays into Po by emitting an alpha-particle with half life of 4 days. A sample contains 6.4 × 10^(10) atoms of Rn. After 12 days, the number of atoms of Rn left in the sample will be a) 3.2 × 10^(10) b) 0.53 ×10^(10) c) 2.1 × 10^(10) d) 0.8 × 10^(10)? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Rn decays into Po by emitting an alpha-particle with half life of 4 days. A sample contains 6.4 × 10^(10) atoms of Rn. After 12 days, the number of atoms of Rn left in the sample will be a) 3.2 × 10^(10) b) 0.53 ×10^(10) c) 2.1 × 10^(10) d) 0.8 × 10^(10)?.
Solutions for Rn decays into Po by emitting an alpha-particle with half life of 4 days. A sample contains 6.4 × 10^(10) atoms of Rn. After 12 days, the number of atoms of Rn left in the sample will be a) 3.2 × 10^(10) b) 0.53 ×10^(10) c) 2.1 × 10^(10) d) 0.8 × 10^(10)? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of Rn decays into Po by emitting an alpha-particle with half life of 4 days. A sample contains 6.4 × 10^(10) atoms of Rn. After 12 days, the number of atoms of Rn left in the sample will be a) 3.2 × 10^(10) b) 0.53 ×10^(10) c) 2.1 × 10^(10) d) 0.8 × 10^(10)? defined & explained in the simplest way possible. Besides giving the explanation of Rn decays into Po by emitting an alpha-particle with half life of 4 days. A sample contains 6.4 × 10^(10) atoms of Rn. After 12 days, the number of atoms of Rn left in the sample will be a) 3.2 × 10^(10) b) 0.53 ×10^(10) c) 2.1 × 10^(10) d) 0.8 × 10^(10)?, a detailed solution for Rn decays into Po by emitting an alpha-particle with half life of 4 days. A sample contains 6.4 × 10^(10) atoms of Rn. After 12 days, the number of atoms of Rn left in the sample will be a) 3.2 × 10^(10) b) 0.53 ×10^(10) c) 2.1 × 10^(10) d) 0.8 × 10^(10)? has been provided alongside types of Rn decays into Po by emitting an alpha-particle with half life of 4 days. A sample contains 6.4 × 10^(10) atoms of Rn. After 12 days, the number of atoms of Rn left in the sample will be a) 3.2 × 10^(10) b) 0.53 ×10^(10) c) 2.1 × 10^(10) d) 0.8 × 10^(10)? theory, EduRev gives you an ample number of questions to practice Rn decays into Po by emitting an alpha-particle with half life of 4 days. A sample contains 6.4 × 10^(10) atoms of Rn. After 12 days, the number of atoms of Rn left in the sample will be a) 3.2 × 10^(10) b) 0.53 ×10^(10) c) 2.1 × 10^(10) d) 0.8 × 10^(10)? tests, examples and also practice JEE tests.
Explore Courses for JEE exam

Top Courses for JEE

Explore Courses
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev