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If radius of O(2) molecule =40A. T=27^(@...

If radius of `O_(2)` molecule `=40A`. `T=27^(@)C` annd P=1atm. Find the time of relaxation.

A

`10^(-10)sec`

B

`10^(-12)sec`

C

`10^(-14)sec`

D

`10^(-8)sec`

Text Solution

Verified by Experts

The correct Answer is:
B
`tau=(lamda)/(V_(rms))=(1)/(sqrt(2)pind^(2))(sqrt(m_(o)))/(sqrt(2RT))`
Now `n=(N)/(V)=(muN_(a))/(V)`
`PV=muRT`
`(mu)/(V)=(P)/(RT)`
so `n=(P)/(RT)xxN_(a)`
`tau=(sqrt(m_(o))RT)/(sqrt(2)mu.PN_(a)d^(2)sqrt(3RT))`
`tau=(sqrt(m_(o)RT))/(sqrt(6)xx3.14xx10^(5)xx6.02xx10^(23)xx16xx10^(-18))`
`=(sqrt(32xx3xx8.3xx10^(-1)))/(sqrt(6)xx3.14xx10^(5)xx6.02xx10^(23)xx16xx10^(-18))`
`=(sqrt(96xx0.83))/(sqrt(6)xx3.14xx6.02xx16xx10^(10))`
`=(4xxsqrt(0.83))/(3.14xx6.02xx16xx10^(10))`
`=(0.9)/(3.14xx6xx4xx10^(10))`
`=0.01xx10^(-10)`
`=10^(-12)` sec
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