Answer:
An alkali metal present in period 2 have larger first ionization energy.
Explanation:
Ionization energy:
The amount of energy required to remove the electron from the atom is called ionization energy.
Trend along period:
As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required.
Trend along group:
As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.
As the size of atom increases the ionization energy from top to bottom also decreases because it becomes easier to remove the electron because of less nuclear attraction and as more electrons are added the outer electrons becomes more shielded and away from nucleus. Thus alkali metal present in period 2 have larger ionization energy because of more nuclear attraction as compared to the alkali metal present in period 4.
Answer:
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Answer:
Explanation:
A function is a relation in which each input has only one output. In the relation , y is a function of x, because for each input x (1, 2, 3, or 0), there is only one output y. x is not a function of y, because the input y = 3 has multiple outputs: x = 1 and x = 2.
Hope That Helps!!!
Brainliest?
1) D. An element
2) D. Suspension
3) A. Iodine
4) A. Atoms of the same element
Number of photons can be calculated by dividing the needed energy by the energy per photon.
The minimum energy needed is given as 2 x 10^-17 joules
Energy per photon = hc / lambda where h is planck's constant, c is the speed of light and lambda is the wavelength
Energy per photon = (<span>6.626 x 10^-34 x 3 x 10^8) / (475 x 10^-9)
= 4.18 x 10^-19 J
number of photons = (2 x 10^-17) / (4.18 x 10^-19)
= 47.79 photons which is approximately 48 photons</span>
