MCQ
If radius of $O _{2}$ molecule $=40 \mathring A, T =27^{\circ} C$ and $P =1 \,atm .$ Find the time of relaxation.
  • A
    $10^{-10} \;sec$
  • $10^{-12}\; sec$
  • C
    $10^{-14}\; sec$
  • D
    $10^{-8}\; sec$

Answer

Correct option: B.
$10^{-12}\; sec$
b
As we know that,

$\tau=\frac{\lambda}{V_{ rms }}=\frac{1}{\sqrt{2} \pi nd ^{2}} \frac{\sqrt{ m _{0}}}{\sqrt{3 RT }}$

Now,

$n =\frac{ N }{ V }=\frac{\mu N _{ a }}{ V }$

And,

$PV =\mu RT$

$\frac{\mu}{ V }=\frac{ P }{ RT }$

So,

$n =\frac{ P }{ RT } \times N _{ a }$

Therefore,

$\tau=\frac{\sqrt{ m _{0} RT }}{\sqrt{2} \pi PN _{ a } d ^{2} \sqrt{3 RT }}$

Substitute the values.

$\tau=\frac{\sqrt{(32)\left(3 \times 8.3 \times 10^{-1}\right)}}{\sqrt{2}(3.14)\left(10^{5}\right)\left(6.02 \times 10^{23}\right)\left(40 \times 10^{-10}\right)^{2}}$

$=0.01 \times 10^{-10}$

$=10^{-12} sec$

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