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Mandarinka [93]
3 years ago
15

Consider these two cases. Case 1: An electron jumps from energy level 5 to energy level 2 in an atom. Case 2: An electron jumps

from energy level 5 to energy level 1 in an atom. For case 1, what happens when an electron jumps from energy level 5 to energy level 2 in an atom? A photon is emitted by the atom. A photon is absorbed by the atom. A proton is absorbed by the atom. A proton is emitted by the atom. Assuming that both cases describe Hydrogen‑like atoms with one electron, for which case is more energy emitted or absorbed? More energy is emitted or absorbed for case 1. It is impossible to tell. The energy is the same for both cases. More energy is emitted or absorbed for case 2.
Chemistry
1 answer:
xenn [34]3 years ago
8 0

Answer:

1.  A photon is emitted by the atom.

2. More energy is emitted or absorbed for case 2

Explanation:

The Bohr's model of the atom was based on quantum mechanics idea. He suggested that electrons move in specific spherical orbits round the nucleus.

The claims of his model emphasized on the premise that electrons only move in permissible orbits or energy levels round the nucleus.

The ground state is the lowest energy state available where n= 1.

The excited state is energy levels of n = 2,3,4 .....

Problem 1

For an electron to return to the ground state from a higher energy level, i.e from  5 to 2 in case 1, energy is emitted in form of photons.

Problem 2

More energy is emitted or absorbed when an electron moves from a higher energy level to a more lower one or jumps from a lower energy level to a more higher one.

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I'm stuck on this assignment, +20 PTS and BRAINLIEST for step by step:
dolphi86 [110]

Answer:

2. 2.74 L

3. 488 K

4. 7.47 L

5. 38.6 L

6. 2.85 mol

7. 319 K

8. 3.43 kPa

Explanation:

Generally speaking, you want to convert units to SI units, but in this case, we are working with ratios.  This makes up for using the units that wouldn't appropriate elsewhere.

2.  Use the equation P₁V₁ = P₂V₂.  Solve for V₂.  

(3.05 L)(870 kPa) = (969 kPa)(V₂)  

V₂ = 2.74 L.  

3.  Use the equation V₁/T₁ = V₂/T₂. Solve for T₂.  

(3.32 L)/(360 K) = (4.50 L)/(T₂)  

T₂ = 488 K.

4.  Do the same as above, but for V₂.  

(5.10 L)/(-56°C) = V₂/(-82°C)  

V₂ = 7.47 L

5.  Use the equation V₁/n₁ = V₂/n₂.  Solve for V₂.  

(37.2 L)/(0.750 mol) = (V₂)/(0.778 mol)  

V₂ = 38.6 L

6.  Do the same as above, but for n₂.

(86.0 L)/(2.65 mol) = (92.5 L)(n₂)  

n₂ = 2.85 mol

7.  Use the equation P₁/T₁ = P₂/T₂.  Solve for T₂.  

(3.00 atm)/(390 K) = (2.45 atm)/(T₂)  

T₂ = 319 K

8.  Do the same as above, but for P₂.  

In this specific case, however you will need to convert units.  Since both temperatures don't have the same sign, the ratio won't come out right.  Convert to Kelvin.  Add 273.15 to the temperature in Celsius to convert to Kelvin -12.3°C = 260.85 K  25°C = 298.15 K.

(3.00 kPa)/(260.85 K) = P₂/(298.15 K)

P₂ = 3.43 kPa

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