IIT JAM Exam > IIT JAM Questions > 24Na decays to one-fourth of its initial amou...
Start Learning for Free
24Na decays to one-fourth of its initial amount in 29.8 hours. Its decay constant is_________ hour–1 (rounded up to four decimal places).
Correct answer is between '0.0460,0.0470'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
24Na decays to one-fourth of its initial amount in 29.8 hours. Its dec...
The decay constant (λ) can be determined using the formula:
N(t) = N(0) * e^(-λt)
where N(t) is the amount of the substance at time t, N(0) is the initial amount of the substance, e is the base of the natural logarithm (approximately 2.718), λ is the decay constant, and t is the time.
In this case, we are given that 24Na decays to one-fourth of its initial amount in 29.8 hours. This means that at time t = 29.8 hours, N(t) = N(0)/4.
Substituting these values into the formula, we get:
N(0)/4 = N(0) * e^(-λ * 29.8)
Dividing both sides of the equation by N(0) and rearranging, we have:
1/4 = e^(-λ * 29.8)
Taking the natural logarithm of both sides, we get:
ln(1/4) = -λ * 29.8
Simplifying, we have:
-λ = ln(1/4) / 29.8
Finally, dividing both sides by -1, we find:
λ = -ln(1/4) / 29.8
Evaluating this expression, we find:
λ ≈ 0.0231 hour^-1
Therefore, the decay constant for 24Na is approximately 0.0231 hour^-1.
N(t) = N(0) * e^(-λt)
where N(t) is the amount of the substance at time t, N(0) is the initial amount of the substance, e is the base of the natural logarithm (approximately 2.718), λ is the decay constant, and t is the time.
In this case, we are given that 24Na decays to one-fourth of its initial amount in 29.8 hours. This means that at time t = 29.8 hours, N(t) = N(0)/4.
Substituting these values into the formula, we get:
N(0)/4 = N(0) * e^(-λ * 29.8)
Dividing both sides of the equation by N(0) and rearranging, we have:
1/4 = e^(-λ * 29.8)
Taking the natural logarithm of both sides, we get:
ln(1/4) = -λ * 29.8
Simplifying, we have:
-λ = ln(1/4) / 29.8
Finally, dividing both sides by -1, we find:
λ = -ln(1/4) / 29.8
Evaluating this expression, we find:
λ ≈ 0.0231 hour^-1
Therefore, the decay constant for 24Na is approximately 0.0231 hour^-1.
Free Test
FREE
| Start Free Test |
Community Answer
24Na decays to one-fourth of its initial amount in 29.8 hours. Its dec...
A0/At=2^n
A0/A0/4=2^n
4=2^n
n=2
n×t1/2=total time
2×t1/2=29.8
t1/2=16.9
dissociation constant =0 .693/16.9. =0.0410
A0/A0/4=2^n
4=2^n
n=2
n×t1/2=total time
2×t1/2=29.8
t1/2=16.9
dissociation constant =0 .693/16.9. =0.0410
|
Explore Courses for IIT JAM exam
|
|
Similar IIT JAM Doubts
24Na decays to one-fourth of its initial amount in 29.8 hours. Its decay constant is_________ hour–1 (rounded up to four decimal places).Correct answer is between '0.0460,0.0470'. Can you explain this answer?
Question Description
24Na decays to one-fourth of its initial amount in 29.8 hours. Its decay constant is_________ hour–1 (rounded up to four decimal places).Correct answer is between '0.0460,0.0470'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about 24Na decays to one-fourth of its initial amount in 29.8 hours. Its decay constant is_________ hour–1 (rounded up to four decimal places).Correct answer is between '0.0460,0.0470'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 24Na decays to one-fourth of its initial amount in 29.8 hours. Its decay constant is_________ hour–1 (rounded up to four decimal places).Correct answer is between '0.0460,0.0470'. Can you explain this answer?.
24Na decays to one-fourth of its initial amount in 29.8 hours. Its decay constant is_________ hour–1 (rounded up to four decimal places).Correct answer is between '0.0460,0.0470'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about 24Na decays to one-fourth of its initial amount in 29.8 hours. Its decay constant is_________ hour–1 (rounded up to four decimal places).Correct answer is between '0.0460,0.0470'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 24Na decays to one-fourth of its initial amount in 29.8 hours. Its decay constant is_________ hour–1 (rounded up to four decimal places).Correct answer is between '0.0460,0.0470'. Can you explain this answer?.
Solutions for 24Na decays to one-fourth of its initial amount in 29.8 hours. Its decay constant is_________ hour–1 (rounded up to four decimal places).Correct answer is between '0.0460,0.0470'. Can you explain this answer? in English & in Hindi are available as part of our courses for IIT JAM.
Download more important topics, notes, lectures and mock test series for IIT JAM Exam by signing up for free.
Here you can find the meaning of 24Na decays to one-fourth of its initial amount in 29.8 hours. Its decay constant is_________ hour–1 (rounded up to four decimal places).Correct answer is between '0.0460,0.0470'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
24Na decays to one-fourth of its initial amount in 29.8 hours. Its decay constant is_________ hour–1 (rounded up to four decimal places).Correct answer is between '0.0460,0.0470'. Can you explain this answer?, a detailed solution for 24Na decays to one-fourth of its initial amount in 29.8 hours. Its decay constant is_________ hour–1 (rounded up to four decimal places).Correct answer is between '0.0460,0.0470'. Can you explain this answer? has been provided alongside types of 24Na decays to one-fourth of its initial amount in 29.8 hours. Its decay constant is_________ hour–1 (rounded up to four decimal places).Correct answer is between '0.0460,0.0470'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice 24Na decays to one-fourth of its initial amount in 29.8 hours. Its decay constant is_________ hour–1 (rounded up to four decimal places).Correct answer is between '0.0460,0.0470'. Can you explain this answer? tests, examples and also practice IIT JAM tests.
|
|
Explore Courses for IIT JAM exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup