Question

In a beryllium atom, if ${a_0} $ be the radius of the first orbit, then the radius of the second will be in general:
A) $n{a_0}$
B) ${a_0}$
C) ${n^2}{a_0}$
D) $\dfrac{{{a_0}}}{{{n^2}}}$

Answer 1

Hint: Beryllium is the fourth element with a total of four electrons. In writing the electron configuration for beryllium the first two electrons will go in the $1s$ orbital. Since $1s$ can only hold two electrons the remaining two electrons for Beryllium go in the $2s$ orbital. Therefore the Beryllium electron configuration will be $1{s^2}2{s^2} $.

Complete step by step solution:
For the ground status of all atoms in a periodic table, radii were obtained that corresponded to the key limit in the radial distribution functions. The relativistic wave functions used were Dirac equations solutions. In order to construct the potential, Slater method was used to use exchange and latter method to correct the self-interaction.

As measurements of the size of each orbital, radiuses corresponding to these limits are used. In the first, second and third set of transitional components, the relationship between orbital radiuses of the electrons of valence is seen as plots.

Since $F$ is directly proportional to ${n^2} $, we get,
$r = \dfrac{{{n^2}{h^2}{\varepsilon _0}}} {{\pi Mz {e^2}}} $
Where, $r$ is the radius of the given orbit, $n$ is the number of Bohr’s orbit, ${\varepsilon _0}$and$\pi $ are constants, $m$ is the mass, $Z$ is the atomic number and $e$ is the charge on proton.
For,
$n = 1$
${r_1} = \dfrac{{{h^2}{\varepsilon _0}}} {{\pi mZ {e^2}}} = {a_0} $
for, ${n^ {th}} $orbit,
${r_n} = \dfrac{{{n^2}{h^2}{\varepsilon _0}}} {{\pi mZ {e^2}}} $
${r_n} = {a_0} {n^2} $
Therefore, the radius of the second will be in general ${a_0} {n^2} $.

The correct answer is option (C).

Note: Just four maximums in the overall charge-density equation exist for the elements below nobelium. The relationship between these maxima and the shell structure of the atom and the influence of external electron shielding is discussed. The effect of relativistic contraction on low angular dynamic orbits is demonstrated.