What is the half-life of a nuclide that loses 38.0% of its mass in 407 hours?
The half-life is the time at which the substance's concentration is reduced by half of its initial amount. The half-life of a nuclide that lost its 38.0% mass is 590 hours, as calculated below.
Understanding Half-Life
Half-life is the time required by a substance to get reduced to half of its initial concentration. In this case, we are determining the half-life of a nuclide that loses 38.0% of its mass in 407 hours.
Given:
Initial quantity of substance (A₀) = 100
Remaining quantity (At) = 62 (100 - 38)
Time elapsed (t) = 407 hours
The rate constant (k) can be calculated using the formula:
ln(At ÷ A₀) = -kt
ln(62 ÷ 100) ÷ 407 hours = -k
-0.47803580094 ÷ 407 = -k
k = 0.00117453513
Now, we can calculate the half-life using the rate constant:
Half-life (t₁/₂) = 0.693 ÷ k
t₁/₂ = 0.693 ÷ 0.00117453513
t₁/₂ ≈ 590 hours
Therefore, the half-life of the nuclide that loses 38.0% of its mass in 407 hours is approximately 590 hours. This means that after 590 hours, the concentration of the substance will be reduced by half of its initial amount.