The half life of radioactive potassium is 1.3 billion years. If 10 gram is present now,how much will be present in 100 years?

Given:
- Half-life of radioactive potassium is 1.3 billion years.
- Initial amount of potassium is 10 grams.

Let's calculate how much will be present in 100 years.

The general formula to calculate the amount of radioactive substance remaining after a certain time is:
N(t) = N_{0} \times \left( \frac{1}{2} \right)^{\frac{t}{T_{\frac{1}{2}}}}
where:
- N(t) is the amount of the substance left after time t,
- N_{0} is the initial amount of the substance,
- t is the time that has passed,
- T_{\frac{1}{2}} is the half-life of the substance.

Given:
- Initial amount N_{0} = 10 grams
- Time t = 100 years
- Half-life T_{\frac{1}{2}} = 1.3 billion years = 1.3 \times 10^9 years

Plugging these values into the formula:
N(100) = 10 \times \left( \frac{1}{2} \right)^{\frac{100}{1.3 \times 10^{9}}}
N(100) = 10 \times \left( \frac{1}{2} \right)^{7.6923 \times 10^{-8}}
N(100) = 10 \times (0.999999999999923)^{7.6923 \times 10^{-8}}
N(100) \approx 10 \times 0.999999999999923 \approx 9.99999999999923 \text{ grams}

\boxed{9.99999999999923} grams of potassium will be present in 100 years.