Po-210 Ia an alpha emitter with a half-life of 138 days. how many grams of Po-210 remain after 552 days if the sample initially contained 5.80g of Po-210?
Suppose that you start with 1000000 atoms of a particular radioactive isotope. how many half-lives woruld be required to reduce the number of undecayed atoms to fewer than 1000?
#1.
k = 0.693/t1/2
Substitute k in the below equation.
ln(No/N) = kt
No = 5.80g
N = ?
K = from above
t = 552 days
Solve for N
For #2, you want 1,000,000/2^n = 1,000
Solve for n. The easy way to do that is to solve for 2^n and then n.
0.3625
To calculate the remaining grams of Po-210 after 552 days, we need to use the equation for radioactive decay:
N = N₀ * (1/2)^(t/h)
Where:
N = Remaining amount of substance
N₀ = Initial amount of substance
t = Time passed
h = Half-life of the substance
Let's substitute the given values into the equation:
N = 5.80g * (1/2)^(552/138)
First, we need to find the value of (1/2)^(552/138):
(1/2)^(552/138) = 0.0625
Now, we can substitute this value back into the original equation:
N = 5.80g * 0.0625
N = 0.3625g
Therefore, after 552 days, there will be 0.3625 grams of Po-210 remaining from the initial 5.80 grams.