the half-life of radon-222 is 3.823 days. what was the original mass if 0.050g remains after 7.646 days?
To find the original mass of radon-222, we can use the formula for exponential decay:
M(t) = M₀ * (1/2)^(t/h)
Where:
- M(t) is the remaining mass at time t
- M₀ is the original mass
- t is the elapsed time
- h is the half-life of the substance
In this case, we are given:
M(t) = 0.050g (remaining mass after 7.646 days)
t = 7.646 days
h = 3.823 days (half-life)
Plugging in these values, we get:
0.050g = M₀ * (1/2)^(7.646/3.823)
To solve for M₀, we need to isolate it on one side of the equation. In this case, we can divide both sides of the equation by (1/2)^(7.646/3.823), which will cancel it on the right side:
M₀ = (0.050g) / (1/2)^(7.646/3.823)
Simplifying the right side gives us:
M₀ = (0.050g) / 2^(7.646/3.823)
Calculating the value of 2^(7.646/3.823) gives approximately 3.9998.
Therefore, the original mass (M₀) is:
M₀ ≈ (0.050g) / 3.9998
M₀ ≈ 0.0125g
So, the original mass of radon-222 was approximately 0.0125 grams.
236
k= 0.693/t1/2
Then substitute k into the equation below.
ln(No/N) = kt
Solve for No.
N = 0.05g
k = from above.
t = 7.646 days.