A radioactive sample contains two different types of radioactive nuclei: A, with half-life 5.0 days, and B, with half-life 30.0 days. Initially, the decay rate of the A-type nucleus is 64 times that of the B-type nucleus. When will their decay rates be equal

t = 39.42 days

Explanation:

To do this, we need to use the general expression for decay rate:

R = R₀ exp(-λt)    (1)

According to the problem, decay rate of A initially is 64 times rate of B, so we can say the following:

A₀ = 64B₀   (2)

The value of λ can be determined using:

λ = ln2 / t(1/2)  (3)

Let's calculate first the value of λ for each nuclei:

For A: λ₁ = ln2 / 5 = 0.1286 day⁻¹

For B: λ₂ = ln2 / 30 = 0.0231 day⁻¹

Now, let's write an expression using (1) for each nuclei.

A = A₀ exp(-0.1286t)

A = 64B₀ exp(-0.1286t)   (3)

B = B₀ exp(-0.0231t)    (4)

We want to know the time when A = B , therefore, we can actually equals (3) and (4) and solve for t:

64B₀ exp(-0.1286t) = B₀ exp(-0.0231t)

64 exp(-0.1286t) = exp(-0.0231t)

ln(64 * exp(-0.1286t)) = ln(exp(-0.0231t))

ln64 + ln(exp(-0.1286t))= -0.0231t

4.1589 - 0.1286t = -0.0231t

4.1589 = (0.1286 - 0.0231)t

t = 4.1589 / 0.1055

So, in 39 days, the decay rates of A and B will be the same.

Hope this helps

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.

the answer is 1260

you spend 7 hours in school for 1 school day with 5 school days in 1 school week with 36 school weeks in a school year.

7*5*36= 1260 it's all about multiplying