Answer
(a) net electrical charge of nucleus $1.312\times10^{-17}C$
(b) no of neutron in lead nucleus is $N=126$
(c) no of neutron in lead nucleus is $A=208$
(d) radius of nucleus $r=7.1099\times10^{-15}m$
(e) nuclear density $\rho=2.31289\times10^{17}kg/m^3$
Work Step by Step
$ ^{208}_{82}Pb$
Comparing with $ ^{A}_{Z}X$
we get
$A=208$
$Z=82$
number of neutrons $N=A-Z$
$N=208-82=126$
no of neutron in lead nucleus is $N=126$
(a) net electric charge of the nucleus
no of protons in nucleus $Z=82$
charge on 1 proton=$+e$=$+1.6\times10^{-19}C$
so $Z=82$ proton will have charge $=Z\times +e=82\times 1.6\times10^{-19}C$
so charge on nucleus $=131.2\times10^{-19}C=1.312\times10^{-17}C$
net electrical charge of nucleus $1.312\times10^{-17}C$
{b)
$A=208$
$Z=82$
number of neutrons $N=A-Z$
$N=208-82=126$
no of neutron in lead nucleus is $N=126$
(C) number of nucleons
$ ^{208}_{82}Pb$
Comparing with $ ^{A}_{Z}X$
we get
$A=208$
no of neutron in lead nucleus is $A=208$
(d) radius of the nucleus
$r=(1.2\times10^{-15}m)A^{\frac{1}{3}}$
putting $A=208$
$r=(1.2\times10^{-15}m)(208)^{\frac{1}{3}}$
$r=7.1099\times10^{-15}m$
(e)nuclear density
mass of nucleus $M=Z\times$ mass of proton $+ N\times $mass of nutron
$M=82\times1.672\times10^{-27}kg + 126\times 1.674\times10^{-27}kg$
$M=348.028\times10^{-27}kg$
considering nucleus as spherical
volume $V=\frac{4}{3} \pi r^{3}$
$V=\frac{4}{3} \pi (7.1099\times10^{-15}m)^{3}$
$V=1504.731\times10^{-45}m^3$
density $\rho=\frac{M}{V}$
$\rho=\frac{348.028\times10^{-27}kg}{1504.731\times10^{-45}m^3}$
$\rho=0.231289\times10^{18}kg/m^3$
so nuclear density $\rho=2.31289\times10^{17}kg/m^3$
