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2 years ago

For a hydrogen atom, which electronic transition would result in the emission of a photon with the highest energy?

Chemistry

1 answer:

BartSMP [9]2 years ago

6 0

Answer: A

Explanation:

Considering the order of filling electrons into orbitals, movement from a higher to a lower energy level results in the emission of a photon with energy equal to the energy difference between the two energy levels. However, the energies of different orbitals are close together for high values of n (principal quantum number). Their relative energies may change significantly when they form ions. This implies that energy levels are better separated and have high differences in energy for low values of n. Hence the answer. This means that photons transiting between these Lowe n levels will posses higher photon energy due to larger energy difference between levels.

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Name:_____________________________________________________ Date:___________ Period:_________ 3/23 - 3/27 Assignment 1: Gas Law P

crimeas [40]

Answer:

The answers are;

1. 8.2 liters

2. 1214.84 ml

3. 318.027 K

4. 4.00 l.

Explanation:

1. Boyle's law states that the volume of given mass of gas is inversely proportional to its pressure at constant temperature

that is

P₁·V₁ = P₂·V₂

Where:

P₁ = Initial pressure = 40.0 mm Hg

V₁ = Initial volume = 12.3 liters

P₂ = Final pressure = 60.0 mm Hg

V₂ = Final volume = Required

From P₁·V₁ = P₂·V₂, V₂ is given by

$V_2=\frac{P_1\cdot V_1}{P_2} = \frac{40.0 mm Hg\cdot 12.3 l}{60.0 mm Hg}$ = 8.2 l

The volume reduces to V₂ = 8.2 liters

2. Here Charles law states that

$\frac{T_1}{V_1} =\frac{T_2}{V_2}$

T₁ = Initial temperature = 27.0 °C = 300.15 K

V₁ = Initial volume = 900.0 mL

T₂ = Final temperature = 132.0 °C = 405.15 K

V₂ = Final volume = Required

Therefore $V_2 =\frac{T_2\cdot V_1}{T_1}$ = $\frac{405.15 K\times 900.0 mL}{300.15 K}$ = 1214.84 ml

V₂ = 1214.84 ml

3. Gay-Lussac's Law states that

$\frac{T_1}{P_1} =\frac{T_2}{P_2}$

Where:

P₁ = Initial pressure = 15.0 atmospheres

T₁ = Initial temperature = 25.0 °C = 298.15 K

P₂ = Final pressure = 16.0 atmospheres

T₂ = Final temperature = Required

∴ $T_2 = \frac{T_1\times P_2}{P_1}$

= $\frac{298.15 K\times 16.0atm}{15.0atm}$ = 318.027 K

T₂ = 318.027 K

4. Avogadro's law states that,

Equal volume of all gases at the same temperature and pressure contain equal number of molecules.

Therefore if 5.00 moles of gas occupies 2.00 l volume, then

1 moles will occupy 2.00/5 l volume and

10 moles will occupy 2.00/5 × 10 or 4.00 l volume.

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