Two radioactive material A and B have decay constants `10 lambda` and `lambda`, respectively. If initially they have a the same number of nuclei, then the ratio of the number of nuclei of A to that of B will be `1//e` after a time `1/(n lambda)`, where n is ___________

Explanation

To solve the problem, we need to analyze the decay of two radioactive materials A and B, which have different decay constants. Here's the step-by-step solution:

### Step 1: Understand the decay law
The number of nuclei remaining after a certain time can be described by the radioactive decay law:
\[ N(t) = N_0 e^{-\lambda t} \]
where:
- \( N(t) \) is the number of nuclei remaining at time \( t \),
- \( N_0 \) is the initial number of nuclei,
- \( \lambda \) is the decay constant,
- \( t \) is the time elapsed.

### Step 2: Define the initial conditions
Let the initial number of nuclei for both materials A and B be \( N_0 \). Thus:
- For material A, the decay constant \( \lambda_A = 10\lambda \).
- For material B, the decay constant \( \lambda_B = \lambda \).

### Step 3: Write the expressions for the number of nuclei
Using the decay law, we can express the number of nuclei remaining for both materials at time \( t \):
- For material A:
\[ N_A(t) = N_0 e^{-10\lambda t} \]
- For material B:
\[ N_B(t) = N_0 e^{-\lambda t} \]

### Step 4: Set up the ratio of the number of nuclei
We need to find the ratio of the number of nuclei of A to that of B:
\[ \frac{N_A(t)}{N_B(t)} = \frac{N_0 e^{-10\lambda t}}{N_0 e^{-\lambda t}} \]
The \( N_0 \) cancels out:
\[ \frac{N_A(t)}{N_B(t)} = \frac{e^{-10\lambda t}}{e^{-\lambda t}} = e^{-10\lambda t + \lambda t} = e^{-9\lambda t} \]

### Step 5: Set the ratio equal to \( \frac{1}{e} \)
We are given that this ratio equals \( \frac{1}{e} \):
\[ e^{-9\lambda t} = \frac{1}{e} \]
This can be rewritten as:
\[ e^{-9\lambda t} = e^{-1} \]

### Step 6: Equate the exponents
Since the bases are the same, we can equate the exponents:
\[ -9\lambda t = -1 \]
This simplifies to:
\[ 9\lambda t = 1 \]

### Step 7: Solve for \( t \)
Now, solving for \( t \):
\[ t = \frac{1}{9\lambda} \]

### Step 8: Compare with the given time
We are given that this time is also expressed as \( \frac{1}{n\lambda} \). Thus, we can set:
\[ \frac{1}{9\lambda} = \frac{1}{n\lambda} \]

### Step 9: Solve for \( n \)
From this equation, we can see that:
\[ n = 9 \]

### Final Answer
Thus, the value of \( n \) is:
\[ n = 9 \]

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