Atoms And Nuclei Question 403
Question: A $ ^{7}Li $ target is bombarded with a proton beam current of $ {{10}^{-4}} $ A for 1 hour to produce $ ^{7}Be $ of activity $ 1.8\times 10^{8} $ disintegrations per second. Assuming that one $ ^{7}Be $ radioactive nucleus is produced by bombarding 1000 protons, determine its half-life.
Options:
A) $ 8.6\times 10^{6}s $
B) $ 4.2\times 10^{5}s $
C) $ 3.1\times 10^{5}s $
D) $ 1.1\times 10^{6}s $
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Answer:
Correct Answer: A
Solution:
- The total number of protons bombarded
$ =\frac{it}{\lambda }=\frac{{{10}^{-4}}\times 3600}{1.6\times {{10}^{-19}}}=22.5\times 10^{17} $
Number of $ ^{7}Be $ produced $ N=\frac{22.5\times 10^{17}}{1000}=22.5\times 10^{14} $
We know that activity $ A=\lambda N $ or $ A=( \frac{0.693}{{t_{1/2}}} )N $
$ \therefore ,{t_{1/2}}=0.693\frac{N}{A}=0.693\times \frac{22.5\times 10^{14}}{1.8\times 10^{8}} $ $ =8.63\times 10^{6}s $
