`'m_(1)'` g of non-radioactive isotopes `._(z)X^(A)` are mixed with `'M_(2)'` g of the radioactive isotopes `._(z)X^(A')`. How much will the specific activity decreases ? Half-life of `._(z)X^(A')=T` Take `N_(A)` as Avagardo number.

Explanation

To solve the problem step by step, we need to calculate the specific activity before and after mixing the non-radioactive and radioactive isotopes, and then find the decrease in specific activity.

### Step 1: Define Specific Activity Before Mixing

The specific activity \( A_1 \) of the radioactive isotope \( _{z}X^{A'} \) before mixing can be defined as:
\[ A_1 = \frac{dN}{dt} = \lambda N \]
where \( \lambda \) is the decay constant and \( N \) is the number of radioactive atoms.

The decay constant \( \lambda \) can be related to the half-life \( T \) by:
\[ \lambda = \frac{\ln(2)}{T} \]

The number of radioactive atoms \( N \) can be calculated using the mass \( M_2 \) of the radioactive isotope and its molar mass \( A' \):
\[ N = \frac{M_2}{A'} \times N_A \]
where \( N_A \) is Avogadro's number.

Thus, we can express the specific activity \( A_1 \) as:
\[ A_1 = \frac{\ln(2)}{T} \cdot \frac{M_2}{A'} \cdot N_A \]

### Step 2: Define Specific Activity After Mixing

After mixing \( m_1 \) g of non-radioactive isotopes and \( m_2 \) g of radioactive isotopes, the total mass becomes \( m_1 + m_2 \). The specific activity \( A_2 \) after mixing can be expressed as:
\[ A_2 = \frac{dN_{\text{total}}}{dt} = \lambda N_{\text{total}} \]

The total number of radioactive atoms \( N_{\text{total}} \) remains the same as before mixing, which is still \( N \) (since non-radioactive isotopes do not contribute to activity):
\[ N_{\text{total}} = \frac{M_2}{A'} \times N_A \]

Thus, the specific activity \( A_2 \) can be expressed as:
\[ A_2 = \frac{\ln(2)}{T} \cdot \frac{M_2}{A'} \cdot N_A \cdot \frac{1}{m_1 + m_2} \]

### Step 3: Calculate the Decrease in Specific Activity

Now, we can find the decrease in specific activity:
\[ \Delta A = A_1 - A_2 \]

Substituting the expressions for \( A_1 \) and \( A_2 \):
\[ \Delta A = \left(\frac{\ln(2)}{T} \cdot \frac{M_2}{A'} \cdot N_A\right) - \left(\frac{\ln(2)}{T} \cdot \frac{M_2}{A'} \cdot N_A \cdot \frac{1}{m_1 + m_2}\right) \]

Factoring out the common terms:
\[ \Delta A = \frac{\ln(2)}{T} \cdot \frac{M_2}{A'} \cdot N_A \left(1 - \frac{1}{m_1 + m_2}\right) \]

### Final Expression for Decrease in Specific Activity

Thus, the final expression for the decrease in specific activity is:
\[ \Delta A = \frac{\ln(2)}{T} \cdot \frac{M_2}{A'} \cdot N_A \cdot \left(\frac{m_1}{m_1 + m_2}\right) \]