# Chapter 20 - Questions and Problems - Page 950: 20.27

$4.89\times10^{19}$ atoms.

#### Work Step by Step

Number of atoms at the beginning $N_{0}=5.00\times10^{22}$ Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{4.20\,min}=0.165\,min^{-1}$ Time $t=42.0\,min$ $\ln(\frac{N_{0}}{N})=kt$ where $N$ is the number of atoms remaining. $\implies \ln(\frac{5.00\times10^{22}}{N})=0.165\times42.0=6.93$ Taking the inverse $\ln$ of both sides, we have $\frac{5.00\times10^{22}}{N}=e^{6.93}=1022.5$ Or $N= \frac{5.00\times10^{22}}{1022.5}=4.89\times10^{19}$

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