The electrons in the second shell of this atom have (3) more energy and the same charge compared to the first shell
Further explanation
Bohr's atomic model has shown the energy levels of electrons in the path of the atomic shell
The greater the value of n (the atomic shell, the main quantum number), the greater the energy level
In normal circumstances, electrons fill the shell at the lowest energy level starting from the shell K, L M and then N
When an atom gets energy from outside, the electrons will absorb energy so it moves to higher energy. This situation is called excited
Electrons will return to the original path or a lower energy level because the excited state is unstable. In this condition, the electron will release energy
The electron energy at the nth path can be formulated:
[tex]\rm En=\dfrac{-Rh}{n^2}[/tex]
Rh = constant 2.179.10⁻¹⁸ J
So the electron transfer energy (ΔE)
ΔE = E end - E initial
From the electron transfer available, because the value of the Rh constant is the same, the effect is the value of n (shell) ⇒ 1 / n²
Electron configuration of Be (Beryllium) with atomic number 4
1s² 2s² or [He] 2s²
So in the first shell E = -1(n=1), the second shell E = -1/4
(n=2)
So the energy value in the second shell is greater than the first shell
While the electron charge is still the same(-1) (such as protons with + charges and neutrons with neutral charges / 0)
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