Question

Every chemical element goes through natural exponential decay, which means that over time its atoms fall apart. The speed of each element's decay is described by its half-life, which is the amount of time it takes for the number of radioactive atoms of this element to be reduced by half. The half-life of the isotope dubnium-263 is 29 seconds. A sample of dubnium-263 was first measured to have 1024 atoms. After t seconds, there were only 32 atoms of this isotope remaining. Write an equation in terms of t that models the situation.

Answer 1

Answer:

t = ln (N/N°)/(-0.0239)

Explanation :

The decay law is represented as

N = N°e^-kt

Where N is the final number of atom,

N° is the initial number of atoms

k is the decay constant

t is the half-life.

From the above we have,

N/N° = e^-kt

take ln of both sides

ln (N/N°) = -kt

t = ln(N/N°)/-k

At half life, N/N° = 1/2

Therefore, t = (ln 1/2)/-k

t = -0.693/-k

But t = 29 sec

29 = -0.693/-k

k = 0.0239 s^-1

Therefore,

The formula will be

t = ln (N/N°)/(-0.0239)

Answer 2

Answer:

An equation that models the problem is 1024x(1/2)^t/29=32