JEE Mains · Physics · STD 12 - 13. Nuclei
Substance \(A\) has atomic mass number \(16\) and half life of \(1\) day. Another substance \(B\) has atomic mass number \(32\) and half life of \(\frac{1}{2}\) day. If both \(A\) and \(B\) simultaneously start undergo radio activity at the same time with initial mass \(320\,g\) each, how many total atoms of \(A\) and \(B\) combined would be left after \(2\) days \(.........\times 10^{24}\)
- A \(3.38\)
- B \(6.76\)
- C \(67.6\)
- D \(1.69\)
Answer & Solution
Correct Answer
(A) \(3.38\)
Step-by-step Solution
Detailed explanation
\(\left( N _0\right) A =\frac{320}{16}=20 \text { moles }\) \(\left( N _0\right) B =\frac{320}{32}=10 \text { moles }\) \(N _{ A }=\frac{\left( N _0\right)_{ A }}{(2)^{2 / 1}}=\frac{20}{4}=5\) \(N _{ B }=\frac{\left( N _0\right)_{ B }}{(2)^{2 / 5}}=\frac{10}{2^4}=0.625\) Total…
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