0:00
Hi everyone.
00:01
Now what they have given, so first of all right the given data that is mass of ab206 that is equal to 0 .68 milligram then a of uranium they have provided 238, 238 then then, e of pb, that is 206, the next mass of uranium 238, uranium 238, that is 1 milligram.
00:59
1 milligram.
01:02
Of course, that is nothing but you can say nt, this is nothing but nt.
01:09
Then t half they provided.
01:13
That is 4 .5 into 10x to 9 the next t they ask then next so even lambda you have to calculate now first part is calculate the mass of the mass of 2 38 uranium that needs decrease that needs decay to produce 0 .68 milligram of 206 .00 of 206 length.
02:28
So yam is equal to 0 .68 into 238 into 238 into 238 divided by 206 and that is equal to 163 .744 divided by 206 and nothing but 075 milligram and therefore now one can easily calculate n 0 .0 is equal to what? milligram of uranium 238 plus uranium 238 decay and that is equal to 1 plus 0 .75m.
03:38
And therefore n is equal to 1 .795.
03:50
Then part b, that is t half is equal to .693 upon lambda...