A radioactive nucleus X decays to a nucleus Y with a decay constant `lambda_X=0.1s^-1`, Y further decays to a stable nucleus Z with a decay constant
`lambda_Y=1//30s^-1`. Initially, there are only X nuclei and their number is `N _0=10^20`. Set up the rate equations for the populations of X, Y and Z. The population of Y nucleus as a function of time is given by `N_Y(t)={N _0lambda_X//(lambda_X-lambda_Y)}[exp(-lambda_Yt)-exp(-lambda_Xt)]`.
Find the time at which `N_Y` is maximum and determine the population X and Z at that instant.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Set up the rate equations

1. **For nucleus X:**
The decay of nucleus X to nucleus Y can be described by the rate equation:
\[ \frac{dN_X}{dt} = -\lambda_X N_X \]
where \( \lambda_X = 0.1 \, \text{s}^{-1} \).

2. **For nucleus Y:**
The change in the number of Y nuclei is given by the formation from X and decay to Z:
\[ \frac{dN_Y}{dt} = \lambda_X N_X - \lambda_Y N_Y \]
where \( \lambda_Y = \frac{1}{30} \, \text{s}^{-1} \).

3. **For nucleus Z:**
The increase in Z nuclei is due to the decay of Y:
\[ \frac{dN_Z}{dt} = \lambda_Y N_Y \]

### Step 2: Find the time at which \( N_Y \) is maximum

Given the function for \( N_Y(t) \):
\[ N_Y(t) = \frac{N_0 \lambda_X}{\lambda_X - \lambda_Y} \left( e^{-\lambda_Y t} - e^{-\lambda_X t} \right) \]
To find the time at which \( N_Y \) is maximum, we need to differentiate \( N_Y(t) \) with respect to \( t \) and set it to zero:
\[ \frac{dN_Y}{dt} = 0 \]

Differentiating \( N_Y(t) \):
\[ \frac{dN_Y}{dt} = \frac{N_0 \lambda_X}{\lambda_X - \lambda_Y} \left( -\lambda_Y e^{-\lambda_Y t} + \lambda_X e^{-\lambda_X t} \right) \]
Setting this equal to zero:
\[ -\lambda_Y e^{-\lambda_Y t} + \lambda_X e^{-\lambda_X t} = 0 \]
Rearranging gives:
\[ \lambda_X e^{-\lambda_X t} = \lambda_Y e^{-\lambda_Y t} \]
Dividing both sides by \( e^{-\lambda_Y t} \):
\[ \lambda_X e^{(\lambda_Y - \lambda_X)t} = \lambda_Y \]
Taking the natural logarithm:
\[ (\lambda_Y - \lambda_X)t = \ln\left(\frac{\lambda_Y}{\lambda_X}\right) \]
Thus,
\[ t = \frac{1}{\lambda_Y - \lambda_X} \ln\left(\frac{\lambda_Y}{\lambda_X}\right) \]

### Step 3: Substitute values to find \( t \)

Substituting \( \lambda_X = 0.1 \) and \( \lambda_Y = \frac{1}{30} \):
\[ t = \frac{1}{\frac{1}{30} - 0.1} \ln\left(\frac{\frac{1}{30}}{0.1}\right) \]
Calculating \( \frac{1}{30} - 0.1 = \frac{1}{30} - \frac{3}{30} = -\frac{2}{30} = -\frac{1}{15} \):
\[ t = -15 \ln\left(\frac{1}{3}\right) = 15 \ln(3) \]
Calculating \( 15 \ln(3) \approx 15 \times 1.0986 \approx 16.48 \, \text{s} \).

### Step 4: Determine populations of X and Z at \( t = 16.48 \, \text{s} \)

1. **Population of X:**
\[ N_X(t) = N_0 e^{-\lambda_X t} = 10^{20} e^{-0.1 \times 16.48} \]
\[ N_X(16.48) = 10^{20} e^{-1.648} \approx 10^{20} \times 0.192 = 1.92 \times 10^{19} \]

2. **Population of Y:**
Using the relationship \( N_Y = \frac{\lambda_X}{\lambda_Y} N_X \):
\[ N_Y(16.48) = \frac{0.1}{\frac{1}{30}} \times 1.92 \times 10^{19} = 3 \times 1.92 \times 10^{19} = 5.76 \times 10^{19} \]

3. **Population of Z:**
Using conservation of nuclei:
\[ N_0 = N_X + N_Y + N_Z \Rightarrow N_Z = N_0 - N_X - N_Y \]
\[ N_Z = 10^{20} - 1.92 \times 10^{19} - 5.76 \times 10^{19} = 10^{20} - 7.68 \times 10^{19} = 2.32 \times 10^{19} \]

### Final Results

- The time at which \( N_Y \) is maximum is approximately \( 16.48 \, \text{s} \).
- The populations at that instant are:
- \( N_X \approx 1.92 \times 10^{19} \)
- \( N_Y \approx 5.76 \times 10^{19} \)
- \( N_Z \approx 2.32 \times 10^{19} \)