Nuclear Physics And Radioactivity Question 497
Question: The half-life of a sample of a radioactive substance is 1 hour. If $ 8\times 10^{10} $ atoms are present at $ t=0 $ , then the number of atoms decayed in the duration $ t=2 $ hour to $ t=4 $ hour will be
Options:
A) $ 2\times 10^{10} $
B) $ 1.5\times 10^{10} $
C) Zero
D) Infinity
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ N=N_{0}{{( \frac{1}{2} )}^{\frac{t}{T_{1l2}}}} $ No of atoms at t = 2hr, $ N_{1}=8\times 10^{10}{{( \frac{1}{2} )}^{\frac{2}{1}}}=2\times 10^{10} $ No. of atoms at t = 4hr, $ N_{2}=8\times 10^{10}{{( \frac{1}{2} )}^{\frac{4}{1}}}=\frac{1}{2}\times 10^{10} $
$ \therefore $ No. of atoms decayed in given duration $ =( 2-\frac{1}{2} )\times 10^{10}=1.5\times 10^{10} $
BETA
