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Untitled - Kelly Walsh High School

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Nuclear Chemistry 301<br />

t 1/2 (ln 2)/k (ln 2)/6.33 10 3 min 1 0.693/6.33 10 3 min 1<br />

109.47867 109 min<br />

A sample of iron-55 weighs 3.75 mg. The decay constant for this isotope is<br />

0.26 y 1 . How much iron-55 will remain in the sample after 6.0 years?<br />

We can recopy and label the variables to get:<br />

N 0 3.75 mg Nt ? k 0.26 y 1 t 6.0 y<br />

The only equation containing these four variables is: ln (N t/N 0) kt<br />

We can enter the values into the equation to get:<br />

ln (0.26 y1 ?<br />

a<br />

) (6.0 y) 1.56 (unrounded)<br />

3.75 mg b<br />

e1.56 ?<br />

a<br />

0.210136 (unrounded)<br />

3.75 mg b<br />

? (0.210136) (3.75 mg) 0.78801 0.79 mg<br />

A sample of argon-41 contains 4.25 mg of this isotope after 254 s. The decay constant<br />

for this isotope is 6.33 10 3 min 1 . What mass of argon-41 was originally<br />

in the sample?<br />

We can recopy and label the variables to get:<br />

N 0 ? N t 4.25 mg k 6.33 10 3 min 1 t 254 s<br />

The only equation containing these four variables is: ln (N t/N 0) kt<br />

We can enter the values into the equation to get:<br />

ln (6.33 103 min1 4.25 mg<br />

a b<br />

) (254 s)<br />

?<br />

The time units do not match; therefore, we need to add a conversion step.<br />

ln (6.33 103 min1 ) (254 s)<br />

0.026797 (unrounded)<br />

e0.026797 1 min<br />

a<br />

60s<br />

4.25 mg<br />

a b<br />

0.97355885 (unrounded)<br />

?<br />

? (4.25 mg)/(0.97355885) 4.3654269 4.37 mg<br />

b<br />

mg<br />

a4.25 b<br />

?

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