The half-life of gold-195m is approximately 30.5
seconds.

Step 1 of 3 : Determine a so that A(t)=A0at
describes the amount of gold-195m left after t seconds, where A0
is the amount at time t=0
. Round to six decimal places.

Answer:

Step-by-step explanation:

The formula for radioactive decay is given by:

A(t) = A0 * e^(-kt)

where A0 is the initial amount of the radioactive substance, A(t) is the amount remaining after time t, k is the decay constant, and e is the base of the natural logarithm.

The half-life of gold-195m is given as t1/2 = 30.5 seconds. We know that after one half-life, the amount of the substance remaining is halved, so we can write:

A(t1/2) = A0 / 2

Substituting these values into the decay equation gives:

A(t1/2) = A0 * e^(-k*t1/2) = A0 / 2

Taking the natural logarithm of both sides gives:

ln(A0 / 2) = -k*t1/2

Solving for k, we get:

k = ln(2) / t1/2

Substituting this value of k back into the decay equation gives:

A(t) = A0 * e^(-ln(2)/t1/2 * t)

Simplifying this expression gives:

A(t) = A0 * (1/2)^(t/t1/2) 

Therefore, we can see that a = (1/2)^(1/t1/2) = (1/2)^(1/30.5) which is approximately equal to 0.977530156. Rounded to six decimal places, a is 0.977530.