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WARRIOR [948]
2 years ago
4

A freshly prepared sample of curium-243 undergoes 3312 disintegrations per second. after 12.0 yr, the activity of the sample dec

lines to 2292 disintegrations per second. the half-life of curium-243 is ________ years.
Chemistry
2 answers:
alina1380 [7]2 years ago
6 0
<span>Answer: 22.59 years
</span><span />

<span>Explanation:
</span><span />

<span>1) The law of disintegration decay is expressed as an exponential equation:


</span><span>N = No x e ^ (-kt)</span>
<span /><span>
Where No is the initial number of particles (atoms), k is the disintegration constant, t is the time, and N is the number of atoms remaining after time t has elapsed.
</span><span />

<span>2) By substituting N = 2292, No = 3312, and t = 12 years, you obtain the value of k:
</span><span />

<span>2292 = 3312 x e^(-k×12) => -12k = ln(2292/3312) = -0.6368127 => k = 0.030677
</span><span />

<span>3) Now, you can use the half-time equation (you can derive it from the equation N = No x e ^(-kt) by doing N = No / 2):
</span>
<span /><span /><span>
</span><span>t-half = ln(2)/k = ln(2)/0.030677 = 22.59 years
</span><span>
</span>
BartSMP [9]2 years ago
4 0
Given:
Initial quantity of the radiation = No =  3312 <span>disintegrations per second
</span>Final quantity of radiation = Nt =  2292 <span>disintegrations per second
</span>time = t = 12 years
 
We know that, 
   t(1/2) = t / (㏒1/2 (Nt/ No)
∴ t(1/2) = 12 / (㏒1/2 (2292/3312)
             = 22.59 years.

<span>The half-life of Curium-243 is <u>22.59 years</u></span>

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