In a physics text book I read the following:
$$e/m=1.758820150(44) ×10^{11} \mathrm{C/kg} $$ In this expression, $(44)$ indicates the likely uncertainty in the last two digits, $50$.
How should I understand this uncertainty? Does it mean $\pm 44$ on the last two digits?
Answer
$$e/m=\color{red}{1.758\,820\,1}\color{blue}{50}\,\color{magenta}{(44)}\color{green}{×10^{11}} \ \mathrm{C/kg}=\left(\color{red}{1.758\,820\,1}\color{blue}{50}\color{green}{×10^{11}} \pm \color{red}{0.000\,000\,0}\color{blue}{44}\color{green}{×10^{11}}\right) \ \mathrm{C/kg}$$
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