1.

If the mean free path fer a given molecule is 1.1 xx 10^(6) cm, what is radius of the molecules (Given number of molecules per cm^(3) = 2·69 xx 10^(19)).

Answer»

`4·362 XX 10^(-8)` CM
`2·181 xx 10^(-8)` cm
`8.724xx10^(-8)` cm
None of these.

Solution :Mean free path is `lamda=1/(sqrt2pind^(2))`
`therefored^(2)=1/(sqrt2pinlamda)`
`=1/(1·414 xx 3.142 xx 2·69 xx 10^(19) xx 1.1 xx 10^(-6))`
`rArrd=8.724xx10^(-8)` cm
Thus correct choice is (a).


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