Question

Measurements of the radioactivity of a certain isotope tell you that the decay rate decreases from 8255 decays per minute to 3110 decays per minute over a period of 4.50 days.

What is the half-life (T1/2) of this isotope?

Answer 1

The half-life of the isotope can be calculated using the formula T1/2 = t / (ln(2) * (Nf / Ni - 1)), where t is the time interval, Nf is the final amount, and Ni is the initial amount.

Step-by-step explanation:

The isotope in question has a decay rate that decreases from 8255 decays per minute to 3110 decays per minute over a period of 4.50 days. To find the half-life of the isotope, we can use the formula:

T1/2 = t / (ln(2) * (Nf / Ni - 1))

Where:

• T1/2 is the half-life in the desired units.
• t is the time interval.
• ln(2) is the natural logarithm of 2.
• Nf is the final amount (3110 decays per minute).
• Ni is the initial amount (8255 decays per minute).

Substituting the given values into the formula, we get:

T1/2 = 4.50 days / (ln(2) * (3110 / 8255 - 1))

Calculating this expression gives us the half-life of the isotope.

Answer 2

Answer:half-life (T1/2) of this isotope =

Step-by-step explanation:

The number of nuclei of any radioactive substance at a given time is expressed by

Nt = N0e⁻kt

Nt=decay of material at a time t, =3110 decays per minute

N=decays at t=0, 8255 decays per minute

k=constant

Nt=N0e−kt

3110= 8255 e⁻k(4.50)

3110/ 8255=e−k(4.50)

0.3767 = e−k(4.50)

In 0.3767 = -k (4.50)

0.976=-4.5k

k=0.976/4.5

=0.2159

Also we know that t 1/2= time that it takes half the original material to decay.it is related to the rate constant by

T₁/₂=ln 2 / k

Therefore half-life (T1/2) of this isotope

T₁/₂=ln 2/0.2159

T₁/₂=3.12 days