t=0:
A A A A A A
A A A A A A
t=25.0s:
A A X A A A
A X A A X A
What is the half-life of A?
a. 6.25s
b. 18.8s
c. 25.0s
d. 50.0s
e.60.2s
I know the correct answer is e. 60.2s but I can t figure out how to get this answer.
In the following diagram, A represents a radioactive isotope that decays into a new isotope, X. Each A represents 1 mmol of atoms.?
2018-08-21 10:51 am
回答 (2)
2018-08-21 2:22 pm
At t = 0 s, there are 12 mmol of radioactive atoms.
At t = 25.0 s, there are 9 mmol of radioactive atoms.
Let n be the number of half-lives.
(1/2)ⁿ = 9/12
log(0.5ⁿ) = log(0.75)
n log(0.5) = log(0.75)
n = log(0.75)/log(0.5)
Half-life = (25.0 s) / [log(0.75)/log(0.5)] = (25.0 s) × log(0.5) / log(0.75) = 60.2 s
The answer: e. 60.2 s
At t = 25.0 s, there are 9 mmol of radioactive atoms.
Let n be the number of half-lives.
(1/2)ⁿ = 9/12
log(0.5ⁿ) = log(0.75)
n log(0.5) = log(0.75)
n = log(0.75)/log(0.5)
Half-life = (25.0 s) / [log(0.75)/log(0.5)] = (25.0 s) × log(0.5) / log(0.75) = 60.2 s
The answer: e. 60.2 s
2018-08-21 12:51 pm
.
Decay formua: ━━━━━➤ ln( N / N₀ ) = -kt
Half-life formula: ━━━━➤ t½ = ln(2) / k
Initial amount N₀ = 12 mmol of atoms
After time t = 25s, amount not decayed yet N(t) = 9 mmol of atoms
ln( N / N₀ ) = -kt
ln( 9 /12 ) = -25k
k = -0.04 ln(0.75)
kt½ = ln(2)
t½ = ln(2) / k
t½ = ln( 2 ) / ( -0.04 ln(0.75) )
t½ = 60.2s
━━━━━
Decay formua: ━━━━━➤ ln( N / N₀ ) = -kt
Half-life formula: ━━━━➤ t½ = ln(2) / k
Initial amount N₀ = 12 mmol of atoms
After time t = 25s, amount not decayed yet N(t) = 9 mmol of atoms
ln( N / N₀ ) = -kt
ln( 9 /12 ) = -25k
k = -0.04 ln(0.75)
kt½ = ln(2)
t½ = ln(2) / k
t½ = ln( 2 ) / ( -0.04 ln(0.75) )
t½ = 60.2s
━━━━━
收錄日期: 2021-04-24 01:10:42
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