Electrons in an atom can be classified as core electrons and valence electrons. Valence electrons are those electrons which are present in valence shell and participates in bond formation. While, Core electrons are all remaining electrons which are not present in valence shell, hence not take part in bonding.
Atomic number of Selenium (Se) is 34 hence it has 34 electrons with following electronic configuration;
1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁴
From electronic configuration it is found that the valence shell is 4, and the number of electrons present in valence shell are 6. So,
Core Electrons = Total Electrons - Valence Electrons
Core Electrons = 34 - 6
Core Electrons = 28
Result:
There are 28 core electrons in Selenium.
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
1.Calcium Chloride
2.Lithium Bromide
3.Beryllium Sulfide
4.Lithium Fluoride
5. Potassium hydroselenide
6. Strontium phosphide
7.Barium Chloride
8.Iron Oxide
9.Iron
10.?
11.Copper Nitride
A scale and a ruler. The scale to measure the mass, and a ruler to measure the volume.
