Question
Question: In a beryllium atom, if \({a_0} \) be the radius of the first orbit, then the radius of the second w...
In a beryllium atom, if a0 be the radius of the first orbit, then the radius of the second will be in general:
A) na0
B) a0
C) n2a0
D) n2a0
Solution
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 1s22s2.
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 n2, we get,
r=πMze2n2h2ε0
Where, r is the radius of the given orbit, n is the number of Bohr’s orbit, ε0andπ are constants, m is the mass, Z is the atomic number and e is the charge on proton.
For,
n=1
r1=πmZe2h2ε0=a0
for, nthorbit,
rn=πmZe2n2h2ε0
rn=a0n2
Therefore, the radius of the second will be in general a0n2.
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.