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March 6, 2015

March 6, 2015

Posted by **Derrick** on Monday, July 19, 2010 at 8:17pm.

A. the mass number

B. the atomic number

C. the number of protons

D. the number of electrons

A is the answer.

After how many half-lives are there equal amounts of parent and daughter isotopes.

A 1

B 2

C 3

D 4

The answer is A.

The half life of Ni-28 is six days. What fraction of a sample of this isotope will remain after 18 days?

A 1/2

B 1/4

C 1/8

D 1/16

- Chemistry -
**Derrick**, Monday, July 19, 2010 at 8:20pmSorry I wasn't finished posting, I accidentally pressed enter and it ended up posting. I don't get how to solve the half-life qquestion.

My other questions are...

The half-life of a particular radioactibe isotope is 6 hours. What percent of the daughter isotope would be present after 1 day?

A 50%

B 75%

C 87.5%

D 93.75%

Your assistance will be longed-for.

- Chemistry -
**DrBob222**, Monday, July 19, 2010 at 9:53pmThe half life of Ni-28 is six days. What fraction of a sample of this isotope will remain after 18 days?

A 1/2

B 1/4

C 1/8

D 1/16

You can do this two or three ways.

a. 18 days/6 days = 3 half lives; the sample will be 1/2 after 1 half life, 1/4 after two half lives, and 1/8 after 3 half lives. This works ok for a small number of half lives as long as they are even (no "part of a half live"). But it wouldn't work very well for say 20 half lives--too long to do it one by one.

b. No/N = where No = initial number of atoms, N = final number of atoms, n = # half lives.

No/N = 2^{n}= 2^{3}= 8. Therefore N = No/8 or 1/8 of the initial number. This works even for fractional half lives.

c. The long way but it is just a variation of the second method above.

k = 0.69315/t_{1/2},

k = 0.693/6 hrs = 0.1155, then substitute into

ln(No/N) = kt

ln(1/N) = 0.1155(18)

(1/N) = 8.00

N = (1/8)No.

The half-life of a particular radioactibe isotope is 6 hours. What percent of the daughter isotope would be present after 1 day?

A 50%

B 75%

C 87.5%

D 93.75%

**24 hours/6 hours 4 half lives.**

(No/N) = 2^{4}= 16.

N = (1/16)No

If we started with 100 for No, then we would be left with 100/16 = 6.25 which means that 100-6.25 = 93.75 have decayed and that is 93.75%

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