**1. The problem statement, all variables and given/known data**

If all of the energy from decay of the 40-K in one banana could be captured and converted into electrical energy, how many bananas would be needed to power a 100 W light bulb?

There is 358 mg potassium-K per 100 grams of banana

half life of potassium-40 is 1.248(3)×109 y

**2. Relevant equations**

ln(A/Ao)=-kt

e=mc2

k= (0.693/half-life)

**3. The attempt at a solution**

If a banana weighs 125 grams on average then there is 447.5 grams of potassium-40 in one banana.

(358)(1.25) = 447.5g

– – – – – – – – – – – – – –

k=(0.693/1.248(3)×10^9 y) = 5.55×10^-10

ln(A/Ao)=-kt

k = 5.55×10^-10

Ao = 447.5 g

t=?

**1. The problem statement, all variables and given/known data**

**2. Relevant equations**

**3. The attempt at a solution**

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