# Difference between revisions of "Solutions 3"

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# After 2.455 • 10<sup>5</sup> years it will be 2.299 • 10<sup>8</sup> Bq <sup>238</sup>U left, which is equal to 18.49 kg.<br><br> | # After 2.455 • 10<sup>5</sup> years it will be 2.299 • 10<sup>8</sup> Bq <sup>238</sup>U left, which is equal to 18.49 kg.<br><br> | ||

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## Latest revision as of 10:01, 9 July 2012

# Amount of Radioactive Material (number of nuclei, number of moles, weigth) and the Law of Radioactive Decay

###### Return to Problem Solving Sets

**1:**

- 3.2 • 10
^{8}Bq - 383 Bq
- 2.3 • 10
^{9}Bq - 8.0 • 10
^{4}Bq

**2:**

- 1.77 • 10
^{13}atoms. - 2.00 • 10
^{11}atoms.

**3:**

2.40 g ^{60}Co.

**4:**

- 2.53 • 10
^{5}Bq. - 7.21 years = 7 years and 76.7 days.

**5:**

The mass of ^{99m}Tc is 20.5 ng. After one week 28 half-lives will have past, and the activity will be the following:

D = D_{0} • (½)^{28} = 4 • 10^{7} Bq • (½)^{28} = 0.15 Bq.

All ^{99m}Tc has disintegrated and turned into ^{99}Tc. The number of ^{99}Tc atoms will therefore be the same as the original number of ^{99m}Tc atoms.

**6:**

- 1 g natural Lu contains 0.0259 g
^{176}Lu, with a half-life equal to 3.8 • 10^{10}years, which gives A = 51 Bq

- There are two naturally occurring radioisotopes of Sm,
^{147}Sm and^{148}Sm, which gives A = 127.24 Bq + 0.001 Bq = 127 Bq.

**7: **

3002 kg

**8:**

- 18.5 kg
- After 2.455 • 10
^{5}years it will be 2.299 • 10^{8}Bq^{238}U left, which is equal to 18.49 kg.