Tritium decays by a first order process that has half-life of 12.5years. how many years will it take to reduce the radio activity of a tritium sample to 15% of it's original value.
k = 0.693/half life = ?
Then ln(No/N) = kt
ln(100/85) = kt
plug k in and solve for time in years.
Post your work if you get stuck.
To find out how many years it will take to reduce the radioactivity of a tritium sample to 15% of its original value, we can use the concept of half-life.
Since the half-life of tritium is given as 12.5 years, it means that in every 12.5 years, half of the tritium sample will decay.
We can use the formula for exponential decay:
N(t) = N0 * (1/2)^(t / t1/2),
Where:
N(t) = the remaining amount of tritium at time t
N0 = the initial amount of tritium
t = time in years
t1/2 = half-life of tritium
Let's say the original value of the tritium sample is 100 units. We want to find out the time it takes for the remaining tritium to be 15 units (15% of the original value).
15 = 100 * (1/2)^(t / 12.5)
Dividing both sides by 100:
15/100 = (1/2)^(t / 12.5)
0.15 = (1/2)^(t / 12.5)
Taking the logarithm to base 2 of both sides:
log2(0.15) = (t / 12.5)
Using a calculator:
t / 12.5 = -2.737
Multiplying both sides by 12.5:
t = -2.737 * 12.5
t = -34.21
The negative value doesn't make sense in this context, as time cannot be negative. Therefore, we can ignore the negative sign and take the absolute value:
t = 34.21 years
So, it will take approximately 34.21 years to reduce the radioactivity of the tritium sample to 15% of its original value.
To determine the number of years it will take to reduce the radioactivity of a tritium sample to 15% of its original value, we can use the concept of half-life and the equation for first-order decay.
The half-life of a substance is the time it takes for half of the substance to decay. In this case, the half-life of tritium is given as 12.5 years, which means that after 12.5 years, the radioactivity will decrease by half.
Let's denote the initial radioactivity as R₀ and the radioactivity after t years as R.
The formula for first-order decay is:
R = R₀ * (1/2)^(t / half-life)
We can rearrange this equation to solve for t:
t = half-life * log2(R / R₀)
Since we want to find the time it takes for the radioactivity to reduce to 15% of the original value, R should be 0.15 * R₀.
Substituting these values into the equation:
t = 12.5 years * log2(0.15 * R₀ / R₀)
Simplifying the equation:
t = 12.5 years * log2(0.15)
Using a calculator or a logarithm table, we can evaluate log2(0.15) to find the value. Let's calculate this:
t ≈ 12.5 years * (-2.736)
t ≈ -34.2 years
The negative sign indicates an invalid result, as time cannot be negative. Therefore, there must have been an error in the calculations.
Please double-check the given values and ensure the correct calculation of log2(0.15).