Answer:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 4p²
Explanation:
This atom will likely have 4 electron shells denotation of – 2.8.8.4
Orbitals shells show the probability, in space around the nucleus, where to find an electron. It is important to note that the 3rd shell has an additional d orbital (-in addition to s and p). However, because the d orbital has a higher energy state than the 4s and 4p orbitals, the d orbital only fills up when these latter ones are completely filled. In this case, the 4p does not completely fill (hence we don't see the d orbital in the notation).
Answer:
1.64x10⁻¹⁸ J
Explanation:
By the Bohr model, the electrons surround the nucleus of the atom in shells or levels of energy. Each one has it's energy, and the electron doesn't fall to the nucleus because it can reach another level of energy, and then return to its level.
When the electrons go to another level, it absorbs energy, and then, when return, this energy is released, as a photon (generally as luminous energy). The value of the energy can be calculated by:
E = hc/λ
Where h is the Planck constant (6.626x10⁻³⁴ J.s), c is the light speed (3.00x10⁸ m/s), and λ is the wavelength of the photon.
The wavelength can be calculated by:
1/λ = R*(1/nf² - 1/ni²)
Where R is the Rydberg constant (1.097x10⁷ m⁻¹), nf is the final orbit, and ni the initial orbit. So:
1/λ = 1.097x10⁷ *(1/1² - 1/2²)
1/λ = 8.227x10⁶
λ = 1.215x10⁻⁷ m
So, the energy is:
E = (6.626x10⁻³⁴ * 3.00x10⁸)/(1.215x10⁻⁷)
E = 1.64x10⁻¹⁸ J
