Answer:
5.41 days
Explanation:
To find the number of days you use the decay law for radioactivity:
[tex]N=N_oe^{- \frac{t}{T_{1/2}}}[/tex]
No: initial amount of 139np
T1/2: half-life = 2.35 days
after the decat, the final amount is 0.100% of the initial 239np, that is N=0.1No. Hence, you have:
[tex]N=0.9N_o=N_oe^{-\frac{t}{T_{1/2}}}[/tex]
by applying properties of logarithms you obtain:
[tex]0.1N_o=N_oe^{-\frac{t}{T_{1/2}}}\\\\ln(0.1)=ln(e^{-\frac{t}{T_{1/2}}})=-\frac{t}{T_{1/2}}\\\\t=-ln(0.1)(T_{1/2})=-(-2.302)(2.35)=5.41days[/tex]
hence, the number of days is 5.41 days
Answer:
23.5 days
Explanation:
The equation for radioactive decay is given by
N = N₀exp(-λt). To find the decay constant, λ, we have
λ = -(lnN/N₀)/t here t = half-life of Neptunium-239 = 2.35 days and N/N₀ = 1/2 where N = Quantity of Neptunium-239 after half-life, N₀ = Initial quantity of Neptunium-239
So, λ = -(ln(1/2))/2.35 = 0.294/day.
Now for Neptunium-239 to decay to 0.100 % of its initial value,
N/N₀ × 100 % = 0.100%
N/N₀ = 0.100/100
N/N₀ = 0.001
From the equation for radioactive decay, we find the time it takes for Neptunium-239 to decay to this value. So, making t subject of the formula,
t = -(lnN/N₀)/λ = -(ln(0.001))/0.294/day = 23.5 days
