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3 years ago

How much of a 400g sample remains after 4 years if a radioactive isotope has a half-life of 2 years?

Chemistry

1 answer:

timama [110]3 years ago

3 0

Answer:

100 g

Explanation:

From the question given above, the following data were obtained:

Original amount (N₀) = 400 g

Time (t) = 4 years

Half-life (t½) = 2 years

Amount remaining (N) =?

Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:

Time (t) = 4 years

Half-life (t½) = 2 years

Number of half-lives (n) =?

n = t / t½

n = 4 / 2

n = 2

Thus, 2 half-lives has elapsed.

Finally, we shall determine the amount remaining of the radioactive isotope. This can be obtained as follow:

Original amount (N₀) = 400 g

Number of half-lives (n) = 2

Amount remaining (N) =?

N = 1/2ⁿ × N₀

N = 1/2² × 400

N = 1/4 × 400

N = 0.25 × 400

N = 100 g

Thus, the amount of the radioactive isotope remaing is the 100 g.

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Answer:

First ionization of lithium:

$\text{Li}\;(g)\to \text{Li}^{+} \; (g) + \text{e}^{-}$ .

Second ionization of lithium:

$\text{Li}^{+}\;(g) \to\text{Li}^{2+} \;(g) + \text{e}^{-}$ .

Explanation:

The ionization energy of an element is the energy required to remove the outermost electron from an atom or ion of the element in gaseous state. (Refer to your textbook for a more precise definition.) Some features of the equation:

Start with a gaseous atom (for the first ionization energy only) or a gaseous ion. Write the gaseous state symbol $(g)$ next to any atom or ion in the equation.
The product shall contain one gaseous ion and one electron. The charge on the ion shall be the same as the order of the ionization energy. For the second ionization energy, the ion shall carry a charge of +2.
Charge shall balance on the two sides of the equation.

First Ionization Energy of Li:

The products shall contain a gaseous ion with charge +1 $\text{Li}^{+}\;(g)$ as well as an electron $\text{e}^{-}$ .
Charge shall balance on the two sides. There's no net charge on the product side. Neither shall there be a charge on the reactant side. The only reactant shall be a lithium atom which is both gaseous and neutral: $\text{Li}\;(g)$ .

Hence the equation: $\text{Li}\;(g) \to \text{Li}^{+}\;(g) + \text{e}^{-}$ .

Second Ionization Energy of Li:

The product shall contain a gaseous ion with charge +2: $\text{Li}^{2+}\;(g)$ as well as an electron $\text{e}^{-}$ .
Charge shall balance on the two sides. What's the net charge on the product side? That shall also be the charge on the reactant side. What will be the reactant?