题目

有一根1 , (m)长的柱子，柱效为3600。在一定条件下，两个组分的保留时间分别为12.2 , (min)和12.8 , (min)，计算分离度。要达到完全分离，所需要的柱长。[W_1 = 4 (t_(r1))/(sqrt(n)) = (4 times 12.2)/(sqrt(3600)) = 0.8133][W_2 = 4 (t_(r2))/(sqrt(n)) = (4 times 12.8)/(sqrt(3600)) = 0.8533]分离度：[R = (2 times (12.8 - 12.2))/(0.8533 + 0.8133) = 0.72][L_2 = ( (R_2)/(R_1) )^2 times L_1 = ( (1.5)/(0.72) )^2 times 1 = 4.34 , (m)]

有一根$1 \, \text{m}$长的柱子，柱效为$3600$。在一定条件下，两个组分的保留时间分别为$12.2 \, \text{min}$和$12.8 \, \text{min}$，计算分离度。要达到完全分离，所需要的柱长。

$
W_1 = 4 \frac{t_{r1}}{\sqrt{n}} = \frac{4 \times 12.2}{\sqrt{3600}} = 0.8133
$

$
W_2 = 4 \frac{t_{r2}}{\sqrt{n}} = \frac{4 \times 12.8}{\sqrt{3600}} = 0.8533
$

分离度：
$
R = \frac{2 \times (12.8 - 12.2)}{0.8533 + 0.8133} = 0.72
$

$
L_2 = \left( \frac{R_2}{R_1} \right)^2 \times L_1 = \left( \frac{1.5}{0.72} \right)^2 \times 1 = 4.34 \, \text{m}
$

题目解答

答案

根据题意，$ W_1 = 4 \times \frac{12.2}{60} = 0.8133\,\text{min} $，$ W_2 = 4 \times \frac{12.8}{60} = 0.8533\,\text{min} $。分离度为： \[ R = \frac{2 \times (12.8 - 12.2)}{0.8133 + 0.8533} = \frac{1.2}{1.6666} = 0.72 \] 要达到 $ R = 1.5 $，需： \[ \frac{R_2}{R_1} = \sqrt{\frac{L_2}{L_1}} \implies \frac{1.5}{0.72} = \sqrt{L_2} \implies L_2 = (2.0833)^2 = 4.34\,\text{m} \] 最终结果：分离度 $ R = 0.72 $，所需柱长 $ L = 4.34\,\text{m} $。