Two radioactive `X_(1)` and `X_(2)` have decay constants `10 lambda ` and `lambda` respectively . If initially they have the same number of nuclei, then the ratio of the number of nuclei of `X_(1)` to that of `X_(2)`will be `1//e` after a time .

Explanation

To solve the problem, we need to find the time \( t \) at which the ratio of the number of nuclei of two radioactive substances \( X_1 \) and \( X_2 \) becomes \( \frac{1}{e} \). Here are the steps to derive the solution:

### Step 1: Define the Initial Conditions
Let the initial number of nuclei of both substances be \( N_0 \).

### Step 2: Write the Decay Equations
For substance \( X_1 \) with decay constant \( 10\lambda \):
\[ N_1(t) = N_0 e^{-10\lambda t} \]
For substance \( X_2 \) with decay constant \( \lambda \):
\[ N_2(t) = N_0 e^{-\lambda t} \]

### Step 3: Set Up the Ratio
According to the problem, we want the ratio of the number of nuclei \( N_1 \) to \( N_2 \) to equal \( \frac{1}{e} \) at some time \( t \):
\[ \frac{N_1(t)}{N_2(t)} = \frac{1}{e} \]

### Step 4: Substitute the Decay Equations into the Ratio
Substituting the expressions for \( N_1(t) \) and \( N_2(t) \):
\[ \frac{N_0 e^{-10\lambda t}}{N_0 e^{-\lambda t}} = \frac{1}{e} \]

### Step 5: Simplify the Ratio
The \( N_0 \) cancels out:
\[ \frac{e^{-10\lambda t}}{e^{-\lambda t}} = \frac{1}{e} \]
This simplifies to:
\[ e^{-10\lambda t + \lambda t} = e^{-1} \]
or
\[ e^{-9\lambda t} = e^{-1} \]

### Step 6: Set the Exponents Equal
Since the bases are the same, we can equate the exponents:
\[ -9\lambda t = -1 \]

### Step 7: Solve for Time \( t \)
Dividing both sides by \(-9\lambda\):
\[ t = \frac{1}{9\lambda} \]

### Final Answer
Thus, the time \( t \) at which the ratio of the number of nuclei of \( X_1 \) to that of \( X_2 \) becomes \( \frac{1}{e} \) is:
\[ t = \frac{1}{9\lambda} \]

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