The radius of a calcium atom is approximately 0.23 nm. What is the magnitude of the ionization energy?

The ionization energy is 111.15 eV.

Explanation:

Ionization energy is the measure of amount of energy required to remove an electron from valence shell. So depending upon the shell and the number of electrons to be removed , the ionization energy will vary. Thus, increase in atomic number will lead to decrease in ionization energy as the increase in number of shells will lead to decrease in the attractive force between nucleus and the shell leading to decrease in ionization energy.

Using Bohr atom model,

Here n is the valence shell principle quantum number and Z is the atomic number.

As the radius is given as 0.23 nm, we can determine the ratio of n²/Z. This value can be substituted in the formula for ionization energy.

Ionization energy =

As radius = 0.23 nm, the Z/n² value is

As, the atomic number of Ca is 20, then

I.E =

So the ionization energy is 111.15 eV.

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2

scattering of light by tiny particles in air to produce the colors we see on sky, is dependent on the following factors:

(i) wavelength of light

(ii) size of the scattering particle  

(iii) distance traveled by light through the atmosphere.

according to rayleigh's law of scattering, the amount of scattering is inversely proportional to the fourth power of its wavelength. shorter the wavelength, greater is its scattering. hence violet/blue light is scattered more than red.

sun's light is a composite light ( visible light) with wavelengths in the range 400 nm (violet) to 700 nm (red). as sunlight passes through our atmosphere, the molecules of air scatter light according to their wavelengths. during sunrise and sunset, the sun's position is below the horizon. hence light from the sun travels a greater distance in the atmosphere when compared to its position a noon.

as sunlight enters the earth's atmosphere, blue light is scattered away and at the end of its journey in the sky, only red and orange wavelengths remain in the beam.

hence at sunrise and sunset, the sky appears reddish orange.