<u>Answer:</u> The correct option is A) They have fixed energy values.
<u>Explanation:</u>
Electron is one of the sub-atomic particle present around the nucleus of an atom which is negatively charged.
In an atomic model, it is assumed that the electron revolves around the nucleus in discrete orbits having fixed energy levels.
These electrons when jumping from one energy level to another, some amount of radiation is either emitted or absorbed.
These fixed energy levels are given by the Bohr model and thus, the electrons are quantized.
Hence, the correct option is A) They have fixed energy values.
Answer:
The 1st and 2nd ones on the top
Explanation:
Hope this helps:)
D = m / V
It even gives you the density of gold in the problem. Major hint. Once you know the volume (using V = m / D) then you can calculate the height (thickness) from the equation...
V = L x W x H
Volume = Length x Width x Height
start by converting 200.0 mg into grams
1000 mg = 1 g
200. mg x (1 g / 10^3 mg) = 0.200 g
V = m / D
V = 0.200 g / (19.32 g/cm^3)
V = 0.01035 cm^3
Convert 2.4 ft and 1 ft to cm
2.4 ft x (12 in / 1 ft) x (2.54 cm / 1 in) = 73.15 cm
1 ft = 30.48 cm
Compute the height (thickness)
V = LxWxH
H = V / LW = 0.01035 cm^3 / 73.15 cm / 30.48 cm
H = 4.64 x 10^-6 cm
Convert to nanometers
4.64 x 10^-6 cm x (1 m / 100 cm) x (10^9 nm / 1 m) = 46.4 nm
Knowing the atomic radius of gold, I might have asked my students for the minimum number of gold atoms in this thickness of gold. This would assume that the gold atoms are all in a row. This would give the minimum number of gold atoms.
Atomic radius gold = 174 pm
Diameter = 348 pm
46.4 nm x (1 m / 10^9 nm) x (10^12 pm / 1 m) x (1 Au atom / 248 pm) = 133 atoms of gold
I think that work is being done on the books because they are being moved to their proper location and they will be sorted properly rather than lying on a table. Without lifting or carrying, you could sort the books by their genre or title name on the bookshelf so it will be sorted much more efficiently.
I’m not sure if this is the answer you are looking for but I hope it helps :)