Question

Radon-220 undergoes alpha decay with a half-life of 55.6 s.?
Assume there are 16,000 atoms present initially and calculate how many atoms will be present at 0 s, 55.6 s, 111.2 s, 166.8 s, 222.4 s, and 278.0 s (all multiples of the half-life). Express your answers as integers separated by commas.
Calculate how many atoms are present at 50 s, 100 s, and 200 s (not multiples of the half-life).

Answer 1

The half life of a radioactive isotope refers to the time taken for half of the number of original number of atoms present in the sample to decay.

The equation below gives the number of atoms present at time t

$N=Noe^-kt$

N = Number of atoms present at time t

No = Number of atoms initially present

k = decay constant

t = time taken

Given that;

t1/2 = 0.693/k

where t1/2 = half life

k = 0.693/t1/2

k = 0.693/ 55.6 s

k = 0.0125 s-1

Substituting values;

N = 16,000 e^-0.0125(0)

N = 16,000 atoms

At 50 s

N = 16,000 e^-0.0125(50)

= 8564 atoms

At 100 s

N = 16,000 e^-0.0125(100)

= 4584 atoms

At 200 s

N = 16,000 e^-0.0125(200)

= 1313 atoms

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