Question

If the Cesium Co nucleus had 78 55 Proton's and neutron's; then it's ? radius would be

Answer 1

total  nucleon of Cs = protons + neutrons

total  nucleon of Cs (A) = 55 + 78 = 133

volume of single nucleon =

70 3

where Ro is radius of single nucleon . it is around in the range of $1\times 10^{-15}$ m

volume of Cs nucleon =

TR² 3

where R is radius of Cs nucleon

so the volume of the Cs nucleus = number of Cs nucleon $\times$   volume of single nucleon

$\frac{4}{3}\pi R^{3} = A\times \frac{4}{3}\pi R_{0}^{3}$

$R^{3} =A\times R_{0}^{3}$

$R =A^{\frac{1}{3}}\times R_{0}$

$R =133^{\frac{1}{3}}\times 1\times 10^{-15}$

$R =5.1\times 10^{-15}m$