Plutonium has atomic mass `210` and a decay constant equal to `5.8 xx 10^(-8)s^(-1)`. The number of `alpha`-particles emitted per second by `1 mg` plutonium is
(Avogadro's constant =` 6.0 xx 10^(23)`).

Options

A. `1.7 xx 10^(9)`
B. `1.7 xx 10^(11)`
C. `2.9 xx 10^(11)`
D. `3.4 xx 10^(9)`

Explanation

To find the number of alpha particles emitted per second by 1 mg of plutonium, we can follow these steps:

### Step 1: Calculate the number of atoms in 1 mg of plutonium.
The number of atoms (N) in a given mass can be calculated using the formula:
\[ N = \frac{m}{M} \times N_A \]
where:
- \( m \) = mass of the substance (in kg)
- \( M \) = atomic mass (in kg/mol)
- \( N_A \) = Avogadro's constant (\( 6.0 \times 10^{23} \) atoms/mol)

Given:
- \( m = 1 \text{ mg} = 1 \times 10^{-3} \text{ g} = 1 \times 10^{-6} \text{ kg} \)
- \( M = 210 \text{ g/mol} = 210 \times 10^{-3} \text{ kg/mol} \)
- \( N_A = 6.0 \times 10^{23} \text{ atoms/mol} \)

Now substituting the values:
\[ N = \frac{1 \times 10^{-6}}{210 \times 10^{-3}} \times 6.0 \times 10^{23} \]
\[ N = \frac{1 \times 10^{-6}}{0.210} \times 6.0 \times 10^{23} \]
\[ N = 4.76 \times 10^{18} \text{ atoms} \]

### Step 2: Use the decay constant to find the number of alpha particles emitted per second.
The rate of decay (number of alpha particles emitted per second) can be calculated using the formula:
\[ \frac{dN}{dt} = -\lambda N \]
where:
- \( \lambda \) = decay constant (\( 5.8 \times 10^{-8} \text{ s}^{-1} \))
- \( N \) = number of atoms calculated in Step 1

Substituting the values:
\[ \frac{dN}{dt} = -5.8 \times 10^{-8} \times 4.76 \times 10^{18} \]
Calculating this gives:
\[ \frac{dN}{dt} = -2.76 \times 10^{11} \text{ s}^{-1} \]

### Step 3: Take the absolute value.
Since we are interested in the number of alpha particles emitted, we take the absolute value:
\[ \left| \frac{dN}{dt} \right| = 2.76 \times 10^{11} \text{ s}^{-1} \]

### Conclusion:
The number of alpha particles emitted per second by 1 mg of plutonium is approximately \( 2.76 \times 10^{11} \).