00:02
Hi, in the first support of this question, we need to find out the concentration of conduction electron which can be given by n is equal to row capital n by m.
00:23
And in this case the density is 8 .95 gram per centimeter cube or we can write it as 8 .9 .5 gram per centimeter cube or we can write it as 8 .8.
00:37
0 .95 into 10 to the power minus 3 kilogram per centimeter cube.
00:45
M has a value of 64 atomic mass units and we know one atomic mass unit correspond to 1 .67 into 10 to the power minus 27 kilogram and capital ns 1 that is the number of variance electron.
01:05
So now upon substituting the values n s equal to 8.
01:10
0 .95 into 10 to the power minus 3 into 1 divided by 64 into 1 .67 into 10 to the power minus 27.
01:23
And upon solving this we will get ns equal to 8 .374 into 10 to the power 22 electrons per centimeter cube.
01:41
So that is the required answer for the first subpart of this.
01:45
Question.
01:48
Now the seconds apart we need to find out the mean free time which can be given by tous equal to m by n e square into sigma.
02:06
M is the mass of electron that is 9 .1 into 10 to the power minus 31 kilograms charge of electron is 1 .6 into 10 to the power minus 19 columns.
02:20
Conductivity is is given as 1 .55 into 10 to the power minus 8 oom meter.
02:28
From the first subpart we have n as 8 .374 into 10 to power 22 electrons per 10 meter cube and we know 1 meter cube corresponds to 10 to 6 cm cube...