Atoms And Nuclei Question 376
Question: Let $ {N_{\beta }} $ be the number of p particles emitted by 1 gram of $ Na^{24} $ radioactive nuclei (half-life =15 hrs) in 7.5 hours, $ {N_{\beta }} $ is close to (Avogadro number $ =6.023\times 10^{23}/gmole $ ):
Options:
A) $ 6.2\times 10^{21} $
B) $ 7.5\times 10^{21} $
C) $ 1.25\times 10^{22} $
D) $ 1.75\times 10^{22} $
Show Answer
Answer:
Correct Answer: B
Solution:
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We know that $ {N_{\beta }}=N_{0}(1-{{e}^{-\lambda t}}) $ $ {N_{\beta }}=\frac{6.023\times 10^{23}}{24}[ 1-e\frac{\ell ,n,2}{15}\times 7.5 ] $
on solving we get, $ {N_{\beta }}=7.4\times 10^{21} $
If $ \alpha $ and B are emitted simultaneously.
