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$10^{-8}\; sec$
$\tau=\frac{\lambda}{V_{ rms }}=\frac{1}{\sqrt{2} \pi nd ^{2}} \frac{\sqrt{ m _{0}}}{\sqrt{3 RT }}$
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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