Atomic weight of a radioactive element is `M_(W)` gm. Radioactivity of m gm of its mass is :-
(`N_(A) = ` Avogardro number, `lambda` = decay constant)

Options

A. `N_(A)lambda`
B. `[(N_(A))/(M_(W))m]lambda`
C. `[(N_(A))/(m)]lambda`
D. `[(N_(A))/(m)M_(w)]lambda`

Explanation

To solve the problem regarding the radioactivity of a radioactive element with atomic weight \( M_W \) grams and mass \( m \) grams, we will follow these steps:

### Step 1: Understand the relationship between radioactivity, decay constant, and number of atoms
The radioactivity \( R \) of a sample is given by the formula:
\[ R = \lambda N \]
where:
- \( R \) is the radioactivity,
- \( \lambda \) is the decay constant,
- \( N \) is the number of atoms in the sample.

### Step 2: Calculate the number of atoms \( N \)
To find \( N \), we need to determine how many moles of the radioactive element we have. The number of moles \( n \) can be calculated using:
\[ n = \frac{m}{M_W} \]
where:
- \( m \) is the mass of the sample,
- \( M_W \) is the atomic weight of the element.

Since one mole contains \( N_A \) atoms (Avogadro's number), the total number of atoms \( N \) can be expressed as:
\[ N = n \times N_A = \frac{m}{M_W} \times N_A \]

### Step 3: Substitute \( N \) into the radioactivity equation
Now we substitute \( N \) back into the radioactivity equation:
\[ R = \lambda N = \lambda \left( \frac{m}{M_W} \times N_A \right) \]
This simplifies to:
\[ R = \frac{\lambda N_A m}{M_W} \]

### Conclusion
Thus, the radioactivity \( R \) of \( m \) grams of the radioactive element is given by:
\[ R = \frac{\lambda N_A m}{M_W} \]