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# . If I have 5.7 g of the radioactive isotope Rn-222 (half-life of 3.8 days),?

Determine how much radon will I have after 11.4 days have passed? (Show or explain how you found your answer). THANK YOU

### 4 Answers

- Dr WLv 76 months agoFavorite Answer
memorize and use this equation

.. A(t) = A(o) * (1/2)^(t / half life)

where

.. A(t) = amount remaining after time = "t" has elapsed

.. A(o) = initial amount

.. t = elapsed time

.. half life = ... half life

this problem

.. A(t) = 5.7g * (1/2)^(11.4days / 3.8days) = 0.71g

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- pisgahchemistLv 76 months ago
Half-life problem ....

Some of these problems require some math, and some are ridiculously easy. Your question is of the latter variety.

11.4 days is three half-lives. At the end of each half-life interval half of the sample has decayed.

0.............1 ..............2 ................3 .............. half-lives

5.7g.....2.85g........1.425g.......0.7125g ....... mass remaining

Round to two significant digits: 0.71g of Rn-222 remains

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Radioactive decay follows first-order kinetics and is described by the integrated rate equation:

A = Ao e^(-kt) .....A and Ao are amounts, t is the elapsed time and k is the decay constant (sometimes expressed as λ) which is related to the half-life.

.... k = ln2 / t½ = ln 2 / 3.8 days = 0.184 day⁻¹

A = Ao e^(-kt)

A = 5.7g x e^-(0.184 day⁻¹ x 11.4 day) = 0.71g

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- davidLv 76 months ago
11.4 / 3,8 = 3 .. so 3 half-lives

1/2 ^3 = 1/8 only 1/8 of the Rn remains

1/8 X 5.7g = 0.71 g remains.

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- Roger the MoleLv 76 months ago
5.7 g x (1/2)^(11.4 days / 3.8 days) = 0.71 g remains

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