To tackle this problem, we need to set up the rate equations for the populations of the radioactive nuclei X, Y, and Z, and then analyze the function for the population of Y to find when it reaches its maximum. Let's break this down step by step.

Setting Up the Rate Equations

We start with the decay of nucleus X to nucleus Y and then Y decaying to stable nucleus Z. The decay constants are given as follows:

• Decay constant for X: x = 0.1 s^-1
• Decay constant for Y: y = 1/30 s^-1

Let’s denote:

• N_X = population of nucleus X
• N_Y = population of nucleus Y
• N_Z = population of nucleus Z

The initial population of X is given as N₀ = 10²⁰.

The rate equations can be expressed as:

• For nucleus X:

  dN_X/dt = -xN_X

• For nucleus Y:

  dN_Y/dt = xN_X - yN_Y

• For nucleus Z:

  dN_Z/dt = yN_Y

Population of Y as a Function of Time

The population of Y as a function of time is given by:

N_Y(t) = (N₀ * x / (x - y)) * (e^-yt - e^-xt)

Finding the Maximum Population of Y

To find the time at which N_Y is maximum, we need to take the derivative of N_Y with respect to time and set it to zero:

dN_Y/dt = 0

Using the product and chain rules, we differentiate:

0 = (N₀ * x / (x - y)) * (-y e^-yt + x e^-xt)

Setting the equation to zero gives us:

-y e^-yt + x e^-xt = 0

Rearranging this leads to:

y e^-yt = x e^-xt

Dividing both sides by e^-yt gives:

y = x e^{-(x - y)t}

Taking the natural logarithm of both sides allows us to solve for t:

t = (1/(x - y)) * ln(x/y)

Calculating the Maximum Population of Y

Now, we can substitute the values of x and y:

x = 0.1 s^-1 and y = 1/30 s^-1 ≈ 0.0333 s^-1

Calculating x - y:

x - y = 0.1 - 0.0333 = 0.0667 s^-1

Now substituting into the time equation:

t = (1/0.0667) * ln(0.1/0.0333)

Calculating this gives:

t ≈ 15.03 seconds

Determining Populations of X and Y at Maximum N_Y

Now that we have the time, we can find the populations of X and Y at this instant:

For N_X:

N_X(t) = N₀ e^-xt

Substituting the values:

N_X(15.03) = 10²⁰ e^{-0.1 * 15.03}

Calculating this gives:

N_X(15.03) ≈ 10²⁰ e^-1.503 ≈ 10²⁰ * 0.223 ≈ 2.23 * 10¹⁹

For N_Y:

N_Y(15.03) = (N₀ * x / (x - y)) * (e^-yt - e^-xt)

Substituting the values:

N_Y(15.03) = (10²⁰ * 0.1 / 0.0667) * (e^{-0.0333 * 15.03} - e^{-0.1 * 15.03})

Calculating this gives:

N_Y(15.03) ≈ 1.5 * 10²⁰ * (0.223 - 0.223) ≈ 1.5 * 10²⁰ * 0.223 ≈ 3.34 * 10¹⁹

Summary of Results

At the time when the population of