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?

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

graph slopes whilst gas cools, then it is flat whilst conensation to liquid happens, the coolong liquid then graphs as a slope when it solidifies graph is flat

latent heat

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y