In 29 days, the number of radioactive nuclei decreases to one-sixteenth the number present initially. What is the half-life (in days) of the material?

The half life of the material approximately 7years

Explanation:

Half life is defined as the time taken by a radioactive material to decay to half of its original substance.

Half life t1/2 = ln2/λ

λ is the decay constant.

To get the decay constant , we will use the relationship

N = Noe^-λt

No is the initial value of the substance

N is the final value of the substance after decay

t is the time taken by the substance to decay.

From the formula

N/No = e^-λt ... (1)

If radioactive nuclei decreases to one-sixteenth the number present initially, this means N = 1/16No

N/No = 1/16

t = 29days

Substituting this values into equation 1 we have:

1/16 = e^-29λ

Taking the ln of both sides

ln(1/16) = lne^-29λ

ln(1/16) = -29λ

λ = ln(1/16)/-29

λ = ln0.0625/-29

λ = -2.77/-29

λ = 0.0955

Substituting λ = 0.0955 into the half life formula, we have;

t1/2 = ln2/0.0955

t1/2 = 7.26days

The half life of the materia is 7.26years

In 29 days, the number of radioactive nuclei decreases to one-sixteenth the number present initially. What is the half-life (in days) of the material

Explanation:

The radioactive decay law says

N(t) = No•2^(-t/t½)

Then,

In 29 days

t = 29

The radioactive element has decreased to one-sixteenth of it's original

Then, N(t) / No = 1/16i

Now,

N(t) = No•2^(-t/t½)

N(t) / No = 2^(-t/t½)

1/16 = 2^(-29/t½)

Take In base 2 of both sides

In(1/16) = In•2^(-29/t½)

In(2^-4) = -29/t½

-4In(2) = -29 / t½

Since the ln is in base 2 then, In(2) =1

-4 = -29 / t½

Therefore

t½ = -29 / -4

t½ = 7.25 days

t½ ≈ 7 days

So, the half life of the radioactive element is approximately 7 days

what is the question?

neither accurate nor precise.