In 1908 Rutherford together with H- Geiger measured the rate of emission of `alpha` particles (x) by radium (in the nature this element is represented by a single nuclide `_(88)^(226)Ra`) and found that `1.00 g` of radius emits `x=3.42xx10^(10) alpha`- particle per second. How many helium atoms were formed after decayed radium atom after `83` days?

Answer

To solve the problem, we need to calculate how many helium atoms are formed from the decay of radium after 83 days. We know that each decay of a radium atom emits one alpha particle, which is equivalent to one helium atom.

### Step-by-Step Solution:

1. **Identify the given data:**
- Rate of emission of alpha particles by 1 g of radium: \( x = 3.42 \times 10^{10} \) alpha particles per second.
- Mass of radium: \( 1 \) g.
- Time duration: \( 83 \) days.

2. **Convert the time from days to seconds:**
\[ \text{Total seconds in 83 days} = 83 \text{ days} \times 24 \text{ hours/day} \times 3600 \text{ seconds/hour} \]
\[ = 83 \times 24 \times 3600 = 7,171,200 \text{ seconds} \]

3. **Calculate the total number of alpha particles emitted in 83 days:**
\[ \text{Total alpha particles} = \text{Rate of emission} \times \text{Total time in seconds} \]
\[ = 3.42 \times 10^{10} \text{ particles/second} \times 7,171,200 \text{ seconds} \]
\[ = 2.448 \times 10^{17} \text{ alpha particles} \]

4. **Determine the number of helium atoms formed:**
Since each alpha particle corresponds to one helium atom, the number of helium atoms formed is equal to the total number of alpha particles emitted:
\[ \text{Number of helium atoms} = 2.448 \times 10^{17} \]

### Final Answer:
The number of helium atoms formed after the decay of radium after 83 days is approximately \( 2.45 \times 10^{17} \).

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