A sample of a radioactive element has a mass of 10 g at an instant t=0. The approximate mass of this element in the sample left after two mean lives is

Options

A. 1.35 g
B. 2.50 g
C. 3.70 g
D. 6.30 g

Explanation

To solve the problem of finding the remaining mass of a radioactive element after two mean lives, we can follow these steps:

### Step 1: Understand the given information
We have a radioactive sample with an initial mass (\(m_0\)) of 10 g at time \(t = 0\). We need to determine the mass remaining after two mean lives.

### Step 2: Define the mean life
The mean life (\(\tau\)) of a radioactive substance is the average time that a nucleus of that substance will exist before decaying. It is related to the decay constant (\(\lambda\)) by the formula:
\[ \tau = \frac{1}{\lambda} \]

### Step 3: Calculate the time after two mean lives
After two mean lives, the time (\(t\)) can be expressed as:
\[ t = 2 \tau = \frac{2}{\lambda} \]

### Step 4: Use the decay formula
The number of radioactive nuclei remaining after time \(t\) can be described by the exponential decay formula:
\[ N_t = N_0 e^{-\lambda t} \]
Since mass is proportional to the number of nuclei, we can express the remaining mass (\(m_t\)) as:
\[ m_t = m_0 e^{-\lambda t} \]

### Step 5: Substitute the values
Substituting \(t = \frac{2}{\lambda}\) into the equation:
\[ m_t = m_0 e^{-\lambda \cdot \frac{2}{\lambda}} = m_0 e^{-2} \]
Given that \(m_0 = 10 \, \text{g}\):
\[ m_t = 10 \, e^{-2} \]

### Step 6: Calculate \(e^{-2}\)
The value of \(e\) is approximately 2.718. Therefore:
\[ e^{-2} \approx \frac{1}{e^2} \approx \frac{1}{(2.718)^2} \approx \frac{1}{7.389} \approx 0.1353 \]

### Step 7: Calculate the remaining mass
Now, substituting \(e^{-2}\) back into the equation for \(m_t\):
\[ m_t \approx 10 \times 0.1353 \approx 1.353 \, \text{g} \]

### Step 8: Round the answer
Rounding 1.353 g gives approximately 1.35 g.

### Final Answer
The approximate mass of the radioactive element left after two mean lives is **1.35 g**.

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