The radius of electron's second stationary orbit in Bohr's atom is R. The radius of 3rd orbit will be

Given below are two statements: one is labelled as Assertion $\mathbf{A}$ and the other is labelled as Reason $\mathbf{R}$

Assertion A: The nuclear density of nuclides ${ }_{5}^{10} \mathrm{~B},{ }_{3}^{6} \mathrm{Li},{ }_{26}^{56} \mathrm{Fe},{ }_{10}^{20} \mathrm{Ne}$ and ${ }_{83}^{209} \mathrm{Bi}$ can be arranged as $\rho_{\mathrm{Bi}}^{\mathrm{N}}>\rho_{\mathrm{Fe}}^{\mathrm{N}}>\rho_{\mathrm{Ne}}^{\mathrm{N}}>\rho_{\mathrm{B}}^{\mathrm{N}}>\rho_{\mathrm{Li}}^{\mathrm{N}}$

Reason R: The radius $R$ of nucleus is related to its mass number $A$ as $R=R_{0} A^{1 / 3}$, where $R_{0}$ is a constant.

In the light of the above statements, choose the correct answer from the options given below