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B210has a half life of 5 days. The time taken for seven-eighth of a sample to decay is
- a)10 days
- b)20 days
- c)3.4 days
- d)15 days
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
B210has a half life of 5 days. The time taken for seven-eighth of a sa...
Half-life of Bi210=5 days
∴k= 0.693/(t1/2) =(0.693/5) day−1
Using k=(2.303/t) log (a/a-x)
(where a = a0, (let) ⇒x=7/8 a0, t is time taken in decay and k is rate constant)
We get, t=(2.303×5/0.693)log a0/(1/8)a0
= (2.303×5/0.693) log8=15days
∴k= 0.693/(t1/2) =(0.693/5) day−1
Using k=(2.303/t) log (a/a-x)
(where a = a0, (let) ⇒x=7/8 a0, t is time taken in decay and k is rate constant)
We get, t=(2.303×5/0.693)log a0/(1/8)a0
= (2.303×5/0.693) log8=15days
Most Upvoted Answer
B210has a half life of 5 days. The time taken for seven-eighth of a sa...
Given: Half-life of B210 = 5 days
To find: Time taken for seven-eighth of a sample to decay
Let the initial mass of the sample be 'm'
After one half-life, the mass of the sample will be m/2
After two half-lives, the mass of the sample will be (m/2)/2 = m/4
Similarly, after three half-lives, the mass of the sample will be (m/4)/2 = m/8
After four half-lives, the mass of the sample will be (m/8)/2 = m/16
After five half-lives, the mass of the sample will be (m/16)/2 = m/32
Thus, we can see that after 5 half-lives, the mass of the sample is less than one-eighth of the initial mass.
Therefore, the time taken for seven-eighth of the sample to decay will be less than 5 half-lives.
Let 't' be the time taken for seven-eighth of the sample to decay.
Then, we can write:
(m/8) = m*(1/2)^(t/5)
Simplifying this equation, we get:
1/2^(t/5) = 1/8
2^(t/5) = 8
2^(t/5) = 2^3
t/5 = 3
t = 15 days
Therefore, the correct option is (D) 15 days.
To find: Time taken for seven-eighth of a sample to decay
Let the initial mass of the sample be 'm'
After one half-life, the mass of the sample will be m/2
After two half-lives, the mass of the sample will be (m/2)/2 = m/4
Similarly, after three half-lives, the mass of the sample will be (m/4)/2 = m/8
After four half-lives, the mass of the sample will be (m/8)/2 = m/16
After five half-lives, the mass of the sample will be (m/16)/2 = m/32
Thus, we can see that after 5 half-lives, the mass of the sample is less than one-eighth of the initial mass.
Therefore, the time taken for seven-eighth of the sample to decay will be less than 5 half-lives.
Let 't' be the time taken for seven-eighth of the sample to decay.
Then, we can write:
(m/8) = m*(1/2)^(t/5)
Simplifying this equation, we get:
1/2^(t/5) = 1/8
2^(t/5) = 8
2^(t/5) = 2^3
t/5 = 3
t = 15 days
Therefore, the correct option is (D) 15 days.
Community Answer
B210has a half life of 5 days. The time taken for seven-eighth of a sa...
Given:
Half-life of B210 = 5 days
To find:
Time taken for seven-eighth of a sample to decay
Solution:
Let the initial amount of the sample be N.
After one half-life period of B210, the amount of sample remaining will be N/2.
After two half-life periods of B210, the amount of sample remaining will be (N/2)/2 = N/2^2.
Similarly, after three half-life periods of B210, the amount of sample remaining will be N/2^3.
After n half-life periods of B210, the amount of sample remaining will be N/2^n.
Now, we need to find the value of n for which the remaining amount of sample is seven-eighth of the initial amount (N).
So, we have:
N/2^n = (7/8)N
Simplifying this equation, we get:
2^n = 8/7
n = log2(8/7)
n = 0.5146
The time taken for n half-life periods to pass is given by:
t = n × half-life period
t = 0.5146 × 5
t ≈ 2.57 days
Therefore, the correct option is (D) 15 days, which is the closest to 2.57 × 6 = 15.42 days.
Half-life of B210 = 5 days
To find:
Time taken for seven-eighth of a sample to decay
Solution:
Let the initial amount of the sample be N.
After one half-life period of B210, the amount of sample remaining will be N/2.
After two half-life periods of B210, the amount of sample remaining will be (N/2)/2 = N/2^2.
Similarly, after three half-life periods of B210, the amount of sample remaining will be N/2^3.
After n half-life periods of B210, the amount of sample remaining will be N/2^n.
Now, we need to find the value of n for which the remaining amount of sample is seven-eighth of the initial amount (N).
So, we have:
N/2^n = (7/8)N
Simplifying this equation, we get:
2^n = 8/7
n = log2(8/7)
n = 0.5146
The time taken for n half-life periods to pass is given by:
t = n × half-life period
t = 0.5146 × 5
t ≈ 2.57 days
Therefore, the correct option is (D) 15 days, which is the closest to 2.57 × 6 = 15.42 days.
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B210has a half life of 5 days. The time taken for seven-eighth of a sample to decay isa)10 daysb)20 daysc)3.4 daysd)15 daysCorrect answer is option 'D'. Can you explain this answer?
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