Bombarding sodium-23 with a proton produces nuclide X and a neutron. What is nuclide X?
^{23}_{10}Ne + e^{+}"> 11 23 N a − − β + − − > 10 23 N e + e + answer: neon
</span>The isotope P <span>has a half-life of 14.3 days. If a sample originally contained 1.00 g of P, </span>
<span>how much was left after 43 days?
</span>
0 days >>> 1.00g
14,3 days >>> 1.00g :2 = 0,5g
(14.3+14.3) days >>> 0.5g : 2 = 0.25g
(14.3+14.3+14.3) days >>> 0.25g : 2 = 0.125g
answer: 0.125g
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<span>Identify X in the reaction
below.
<span> U </span><span>+ C </span><span>→ Cf </span><span>+ X
</span>
</span>[tex]^{238}_{92}U ---> ^{251}_{98}Cf+6e^-[/tex]
<span>
answer: 6 electrons (so it's beta</span>⁻)
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<span>A .20 gram sample of C-14 was allowed to decay for 3 half-lives. What mass
<span>of the sample will remain? Carbon-14 has a half life of 5730 </span>
<span>years.
</span>
at the beginning >>> 0.20g
1 half-live >>> 0.20g : 2 = 0.10g
2 half-live >>> 0.10g : 2 = 0.05g
3 half-live >>> 0.05g : 2 = 0.025g
answer: 0.025g
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</span><span>The isotope Cu <span>has a half-life of 30 s. If a sample originally contained 48 mg of Cu, </span>
<span>how much time passed before the amount fell to 3 mg?
</span>
0 s (at the beginning) >>> 48mg
30s >>>>>>>>>>>>>>> 48mg : 2 = 24mg
(30+30)s >>>>>>>>>>> 24mg : 2 = 12mg
(30+30+30)s >>>>>>>> 12mg : 2 = 6mg
(30+30+30+30)s >>>>> 6mg : 2 = 3mg
answer: 30+30+30+30 = 120s
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</span>What radionuclide decays to Fe-56 <span>by beta emission?
</span>
\[^{56}_{26}Fe--^{\beta^-}-->^{56}_{27}Co + e^-\]
answer: ⁵⁶₂₇Co
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The Cf <span>to Cf </span><span>conversion is accompanied by __________.
</span>
Cf to Cf ???? maybe...mistake?
The nuclide produced by bombarding sodium-23 with a proton and yielding a neutron is magnesium-24. Therefore, the answer is A. magnesium-24.
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