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JEE Mains · Physics · STD 12 - Nuclei
  • JEE Mains 2023
  • Medium
  • MCQ

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 \(.........\times 10^{24}\)

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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 \(N =5.625\)

No. of atoms \(=5.625 \times 6.023 \times 10^{23}\)

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