Question

A sample of $Ne$ is originally $10$% by mole in ${Ne}^{20}$ isotopes and remaining are in ${Ne}^{22}$ isotope. In $n$ diffusion steps $25$%, ${Ne}^{20}$ can be achieved. Find $\left(\frac{n}{6}\right)$:
(Given $\log 3=0.48,\log 1.1=0.04$)

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is D 3

Rate of diffussion is given by:

$\frac{{Rate}_1}{{Rate}_2}=\sqrt{\frac{M_2}{M_1}}$

For $M_1=20$ and $M_2=22$ for ${}_{20}Ne$ and ${}_{22}Ne$

$\frac{{Rate}_1}{{Rate}_2}=\sqrt{\frac{22}{20}}=\sqrt{1.1}$

=separation

To obtain 25% ${}_{20}Ne$ let $n$ steps required. An equation that relates the initial and final proportion of an isotope to the number of times the separation process is repeated is:

final purity = (initial purity)(separation)${}^n$

In this case, $0.25=\left(0.10\right)(1.1{)}^{n/2}$,

$\log 0.25=\log 0.10+\frac{n}{2}\log 1.1\frac{n}{2}\log 1.1=1-2\log 5n=19.23$

Thus $n/6=19.23/6=3.2$

therefore option D is correct