Neptunium-239 has a half-life of 2.35 days. How many days must elapse for a sample of 239np to decay to 0.100% of its original quantity?

Question

Neptunium-239 has a half-life of 2.35 days. How many days must elapse for a sample of 239np to decay to 0.100% of its original quantity?

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Dâu 4 years 2021-07-27T10:46:04+00:00 2 Answers 114 views 0

Answers ( )

  1. linhdan
    0
    2021-07-27T10:47:35+00:00
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    Answer:

    5.41 days

    Explanation:

    To find the number of days you use the decay law for radioactivity:

    N=N_oe^{- \frac{t}{T_{1/2}}}

    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:

    N=0.9N_o=N_oe^{-\frac{t}{T_{1/2}}}

    by applying properties of logarithms you  obtain:

    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

    hence, the number of days is 5.41 days

  2. binhan
    0
    2021-07-27T10:47:52+00:00
    Reply

    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

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