Question

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 $\frac1{ 2}$ 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.

• A. $3.38×10^{24 }$
• B. $ 1.69×10^{24 } $
• C. $ 6.76×10^{24 } $
• D. $ 6.76×10^{23} $

Answer 1

The Correct Option is A

For substance A:
\[ (N_0)_A = \frac{320}{16} = 20 \, \text{moles} \] \[ N_A = (N_0)_A \left(\frac{1}{2}\right)^2 = 20 \times \frac{1}{4} = 5 \, \text{moles} \] For substance B:
\[ (N_0)_B = \frac{320}{32} = 10 \, \text{moles} \] \[ N_B = (N_0)_B \left(\frac{1}{2}\right)^4 = 10 \times \frac{1}{16} = 0.625 \, \text{moles} \] Total moles = \(5 + 0.625 = 5.625\). The total number of atoms: \[ 5.625 \times 6.023 \times 10^{23} = 3.38 \times 10^{24} \]