What percentage of sodium-24 with a half life of 15.0 hr would remain after 60hours?
k = 0.693/t1/2. Calculate k and substitute into the equation below.
ln(No/N) = kt
I would start with a convenient number (say 100) for No, calculate N (which is the number remaining after 60 hours). Then (N/No)*100 = percent remaining.
To calculate the percentage of sodium-24 remaining after 60 hours, we can use the formula for radioactive decay:
N(t) = N₀ * (1/2)^(t / t₁/₂)
Where:
N(t) is the amount of substance remaining after time t
N₀ is the initial amount of substance
t₁/₂ is the half-life of the substance
t is the time in hours
In this case, with a half-life of 15.0 hours, we can calculate as follows:
N(60) = N₀ * (1/2)^(60 / 15)
Let's calculate that:
N(60) = N₀ * (1/2)^4
N(60) = N₀ * 1/16
To find the percentage remaining, we need to divide N(60) by N₀ and multiply by 100:
Percentage remaining = (N(60) / N₀) * 100
Percentage remaining = (N₀ * 1/16 / N₀) * 100
Percentage remaining = (1/16) * 100
Percentage remaining = 6.25%
Therefore, approximately 6.25% of the sodium-24 would remain after 60 hours.
To calculate the percentage of sodium-24 that would remain after 60 hours, we need to use the concept of half-life. The half-life of sodium-24 is 15.0 hours, which means that every 15.0 hours, the amount of sodium-24 is reduced by half.
To solve the problem, we can divide the given time (60 hours) by the half-life (15 hours) to determine how many half-lives have passed.
60 hours ÷ 15 hours = 4 half-lives
Since each half-life reduces the amount of sodium-24 by half, after 4 half-lives, the amount would be reduced by a factor of 2 raised to the power of 4.
2^4 = 16
Therefore, the amount of sodium-24 remaining after 60 hours would be 1/16th (or 1/2^4) of the original amount.
To convert this to a percentage, we multiply the fraction by 100:
(1/16) * 100 = 6.25%
Therefore, approximately 6.25% of sodium-24 would remain after 60 hours.