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.
Solution
<p>${n_A} = 20$ moles</p>
<p>${n_B} = 10$ moles</p>
<p>$N = {N_0}{e^{ - \lambda t}}$</p>
<p>${N_A} = (20\,N){e^{ - \left( {{{\ln 2} \over 1} \times 2} \right)}}$</p>
<p>$= {{20\,N} \over 4} = 5\,N$ (N = Avogadro's Number)</p>
<p>${N_B} = 10\,N\,{e^{ - 4\ln 2}}$</p>
<p>$= \left( {{{10\,N} \over {16}}} \right)$</p>
<p>$${N_A} + {N_B} = 5\,N + {{10\,N} \over {16}} = \left( {{{90\,N} \over {16}}} \right) = 3.38 \times {10^{24}}$$</p>
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Bohr's Model of Hydrogen Atom
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