00:01
Solving part a of this problem, as we all know that the radius of nucleus r is equal to 1 .2 into 10 to the power minus 15 meter multiplication a to the power 1 by 3.
00:16
So a is here to 7 to 8 whole to the power 1 by 3.
00:22
Now we also know that volume of nucleus as 4 by 3 pi r cubed.
00:28
So just putting the value of r here, so i can write 4 by 3 multiplication 1 .2 into 10 to the power minus 15 cube multiplication here it will be pi.
00:47
Multiplication 2 to 8 to the power 1 by 3 whole to the power 3.
00:57
On solving it, i get the value as, just look at it carefully, 4 by 3, pi multiplication 1 .2 into 10 to the power minus 15 meter cube multiplication 2 to 8.
01:16
Now the nuclear density is given by rho is equal to m by v is equal to 2 to 2 to 8 multiplication 1 .6 7.
01:29
Into 10 to the power minus 27 by v.
01:34
V is 4x3 pi 1 .2 into 10 to the power minus 15 cube multiplication 2 to 8.
01:43
On solving it, i get the nuclear density of the given element row as 2 .3 into 10 to the power 17 kg per meter cube.
01:57
Now solving part b, so row is equal to m by v, which can be written as am by 4 by 3, pi multiplication 1 .7 into 10 to the power minus 45 multiplication a...