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Question
for the nuclear reaction $^{110}_{47}ag\to^{0}_{- 1}e+^{110}_{48}cd$. starting with 200. g of ag - 110 in a container, how long will it take to decay to only 6.25 g? the half - life of ag - 110 is 252 days. for each half - life we divide by 2. so, in order to reach 6.25 g we have to divide 200. g by 2 1 times. the number of half - lives is 2. each half - life is 252 days so we multiply this by the number of half - lives. the time required is thus 3 days. the total mass in the container is still about 4 g. most of the mass is now 5. a. 200. b. 110 c. 100. d. 50.0 e. 25.0 f. 12.5 g. 6.25 h. ag - 110 i. cd - 110 j. electron k. 756 l. 1 m. 2 n. 3 o. 4 p. 5 q. 6 r. 7 s. 1512 t. 1260 u. 1008 v. 756 w. 504
Step1: Calculate number of half - lives
We know the formula for radioactive decay $N = N_0\times(\frac{1}{2})^n$, where $N$ is the final amount, $N_0$ is the initial amount, and $n$ is the number of half - lives. Given $N_0 = 200$ g and $N=6.25$ g. So, $6.25 = 200\times(\frac{1}{2})^n$. Then $(\frac{1}{2})^n=\frac{6.25}{200}=\frac{1}{32}$. Since $\frac{1}{32}=(\frac{1}{2})^5$, the number of half - lives $n = 5$. So the answer for 1 and 2 is P. 5.
Step2: Calculate time required
The half - life $t_{1/2}=252$ days. The time $t=n\times t_{1/2}$. Substituting $n = 5$ and $t_{1/2}=252$ days, we get $t=5\times252 = 1260$ days. So the answer for 3 is T. 1260.
Step3: Total mass in container
According to the law of conservation of mass in a nuclear reaction (neglecting mass - energy conversion effects in this simple case), the total mass in the container remains the same as the initial mass because mass is conserved. So the total mass is 200 g. The answer for 4 is A. 200.
Step4: Identify most of the mass
In the nuclear reaction $^{110}_{47}Ag
ightarrow^{0}_{ - 1}e+^{110}_{48}Cd$, Ag - 110 decays to Cd - 110 and an electron. The electron has a very small mass compared to the mass of the nucleus. So most of the mass is now Cd - 110. The answer for 5 is I. Cd - 110.
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- P. 5
- P. 5
- T. 1260
- A. 200
- I. Cd - 110