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Aleksandr [31]
3 years ago
14

Explain why a 5eV photon of light cannot be absorbed or emitted by the hydrogen atom

Physics
1 answer:
sweet [91]3 years ago
5 0

Answer:

None of the transitions in the hydrogen atom corresponds to a photon energy of 5eV hence no photon of this energy is absorbed or emitted by the hydrogen atom.

Explanation:

Electrons in a hydrogen atom must be in one of the allowed energy levels. If an electron is in the first energy level, it must have exactly -13.6 eV of energy. If it is in the second energy level, it must have -3.4 eV of energy and so on.

If the electron wants to jump from the first energy level, n = 1, to the second energy level n = 2. The second energy level has higher energy than the first, so to move from n = 1 to n = 2, the electron needs to gain energy. It needs to gain (-3.4) - (-13.6) = 10.2 eV of energy to be excited to the second energy level.

The step from the second energy level to the third is much smaller. It takes only 1.89 eV of energy for this excitation to take place. It takes even less energy to excite electrons in hydrogen from the third energy level to the fourth, and even less from the fourth to the fifth.

None of these transitions in the hydrogen atom corresponds to a photon energy of 5eV hence no photon of this energy is absorbed or emitted by the hydrogen atom.

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The elements selenium and oxygen are both poor electrical conductors. In which section of the periodic table are they
murzikaleks [220]

Answer:

The answer to your question is Nonmetals

Explanation:

Nonmetals   they are bad conductors of heat and electricity except graphite.

Metalloids   they are less conductors of electricity than metals.

Noble gases  they conduct electricity.

Halogens they are not metals and do not conduct electricity.

From this information, we conclude that Oxygen and Selenium are nonmetals.

6 0
3 years ago
Fix any punctuation or capitalization errors below. Click "Submit Answer" if there
asambeis [7]

Answer:slightly broken

Explanation:

8 0
3 years ago
Read 2 more answers
The ΔG of a reaction would be at the minimum when... Group of answer choices the equilibrium constant is equal to 1 (i.e., the r
Blizzard [7]

Answer:

the equilibrium constant is equal to 1 (i.e., the reactant and product concentrations are always equal).

Explanation:

ΔG is a symbol related to Gibbs free energy, which is a physical quantity related to thermodynamics. ΔG refers to the difference between the change in enthalpy (and sometimes entropy) and the temperature of a chemical reaction.

Gibbs free energy is very useful for measuring the work done between the reactants in a reaction. It is calculated using the formula: ΔG = change in enthalpy - (temperature x change in entropy).

The ΔG of a reaction would have a minimum value (zero), if the equilibrium constant is equal to 1 (that is, the concentrations of the reagent and the product are always equal).

5 0
4 years ago
frictional force acting on a 0.5 kg object and a floor is 3 N. What is the coefficient of friction between the object and the fl
ikadub [295]

Answer:

μ  = 0.6

Explanation:

F =  μN

N = mg

F = μmg

3 N =  μ*0.5 kg * 9.8 m/s²

μ = 3/(0.5*9.8) = 0.6

7 0
3 years ago
Ear "popping" is an unpleasant phenomenon sometimes experienced when a change in pressure occurs, for example,in an airplane. If
marshall27 [118]

Answer:

H = 0.00058m = 0.58mm of Mercury

<em><u>h = 160m for popping at 8000m</u></em>

Explanation:

Air density decreases with altitude (increasing height).

At sea level air density = 1.22kg/m^3

while at 3000m above it is approximately 0.8 kg/m^3

ASSUMPTION:

we neglect the change in density across 100m vertically

density of mercury = 13600kg/m ^3

we equate the pressure for mercury and air

AIR(pgh) = Mercury(pgh)

0.8 * g * 100 = 13600 * g * H

H = 0.00058m = 0.58mm of Mercury

At even farther heights, the air density further drops

at 8000m above it is 0.52kg/ m^3  (source:https://www.engineeringtoolbox.com/standard-atmosphere-d_604.html)

***for the ear to pop again, it has to experience same change in pressure as of at 3000m

Change in P for AIR at 3000m = Change in P for AIR at 8000m

ASSUMPTION : gravitational acceleration does not change with altitude

pgh = pgh

0.8 * g * 100 = .5 * g *<em><u>h</u></em>

<em><u>h = 160m</u></em>

3 0
4 years ago
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