^235UF6 and ^238UF6 (molar masses 349 and 352) are separated by effusion. The separation factor per stage is sqrt352/349. The minimum number of stages needed to raise the ^235U/^238U ratio by a factor of 1.05 is nearest to (use ln(352/349) = 0.00856, ln 1.05 = 0.04879)

JEE Advanced levelSingle correct optionDifficulty 3 of 3ig145

$^{235} UF X_{6}$ and $^{238} UF X_{6}$ (molar masses 349 and 352) are separated by effusion. The separation factor per stage is $352/349$ . The minimum number of stages needed to raise the $^{235} U /^{238} U$ ratio by a factor of $1.05$ is nearest to (use $ln (352/349) = 0.00856$ , $ln 1.05 = 0.04879$ )

Per stage the ratio is multiplied by

(352/349)^{1/2}

. After

n

stages:

(352/349)^{n /2} = 1.05

, so

n = \frac{2 ln 1.05}{ln ( 352/349 )} = \frac{2 \times 0.04879}{0.00856} = 11.4

, so

12

stages.