The half-life of radium is `1620 years` and its atomic weight is `226`. The number of atoms that will decay from its `1 g` sample per second will be .

Options

A. `3.6 xx 10^(10)`
B. `3.6 xx 10^(12)`
C. `3.1 xx 10^(15)`
D. `31.1 xx 10^(15)`

Explanation

To solve the problem of determining the number of atoms that will decay from a 1 g sample of radium per second, we will follow these steps:

### Step 1: Calculate the decay constant (λ)
The decay constant (λ) can be calculated using the formula:

\[ \lambda = \frac{0.693}{T_{1/2}} \]

where \(T_{1/2}\) is the half-life of the substance. Given that the half-life of radium is 1620 years, we first need to convert this into seconds.

1. Convert years to days:
\[ 1620 \text{ years} \times 365 \text{ days/year} = 591300 \text{ days} \]

2. Convert days to hours:
\[ 591300 \text{ days} \times 24 \text{ hours/day} = 14191200 \text{ hours} \]

3. Convert hours to seconds:
\[ 14191200 \text{ hours} \times 3600 \text{ seconds/hour} = 51091200000 \text{ seconds} \]

Now substitute \(T_{1/2}\) in seconds into the decay constant formula:

\[ \lambda = \frac{0.693}{51091200000} \approx 1.356 \times 10^{-11} \text{ s}^{-1} \]

### Step 2: Calculate the number of atoms (N) in 1 g of radium
To find the number of atoms in 1 g of radium, we use Avogadro's number and the atomic weight of radium:

\[ N = \frac{N_A}{\text{Atomic Weight}} \]

where \(N_A = 6.023 \times 10^{23} \text{ atoms/mol}\) and the atomic weight of radium is 226 g/mol.

\[ N = \frac{6.023 \times 10^{23}}{226} \approx 2.67 \times 10^{21} \text{ atoms} \]

### Step 3: Calculate the number of decays per second (dn/dt)
Using the radioactive decay law, we can find the rate of decay:

\[ \frac{dn}{dt} = -\lambda N \]

Substituting the values we calculated:

\[ \frac{dn}{dt} = - (1.356 \times 10^{-11}) \times (2.67 \times 10^{21}) \]

Calculating this gives:

\[ \frac{dn}{dt} \approx 3.62 \times 10^{10} \text{ decays/second} \]

### Final Answer
The number of atoms that will decay from the 1 g sample of radium per second is approximately \(3.6 \times 10^{10}\).

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