Question

A radioactive substance decays exponentially. a scientist begins with 190 milligrams of a radioactive substance. after 17 hours, 95 mg of the substance remains. how many milligrams will remain after 21 hours

Answer 1

To calculate how many milligrams will remain after 21 hours, we can use the formula for exponential decay. We are given that after 17 hours, 95 mg of the substance remains. Solving for the remaining amount after 21 hours gives us approximately 88.66 mg.

Step-by-step explanation:

To calculate how many milligrams will remain after 21 hours, we can use the formula for exponential decay:

R = Ro * e^(-kt)

Where:

• R is the remaining amount after time t
• Ro is the initial amount
• k is the decay constant
• t is the time period

We are given that after 17 hours, 95 mg of the substance remains. Plugging these values into the formula, we can solve for k:

95 = 190 * e^(-17k)

Simplifying, we get:

e^(-17k) = 1/2

Taking the natural logarithm (ln) of both sides:

-17k = ln(1/2)

Solving for k, we find k = ln(1/2) / -17.

Now, we can use this value of k to find the remaining amount after 21 hours:

R = 190 * e^(-kt)

R = 190 * e^(-(ln(1/2) / -17) * 21)

Calculating this expression will give us the remaining amount after 21 hours, which is approximately 88.66 mg.

Answer 2

To solve this item, we need write first all the necessary data to solve the problem. Governing principle of this problem is basically exponential decay of a certain substance.

Given:

Let t represents time
N represents mass of substance

Initially,
at t(0), N(0) = 190mg radioactive substance

at t(17), N(17) = 95mg radioactive substance


Required: N(21)

Solution:

We write first the formula for exponential decay, that is

N(t) = N(0)*e^(-k*t) ; k = exponential decay constant

We need first to determine the k, substituting the given we have
95 = 190*e^(-k*17)

Dividing by 190, we get
0.5 = e^(-k*17)

To eliminate the exponential e, we take the natural logs of both sides
ln(0.5) = ln(e^(-k*17))
-0.693 = -k*17

Solving for k,
k = 0.04077

To solve for N(21), we need to substitute k to the original equation
N(21) = N(0)e^(-k*21)
N(21) = 190*e^(-0.04077*21)
N(21) = 80.7 mg

ANSWER: 80.7 mg radioactive substance left after 21 hours