All categories

3 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]3 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.

You might be interested in

Explain how deforestation can cause a disruption to the carbon cycle

TEA [102]

Trees and forest play an important role in balancing the amount of Carbon in the atmosphere through the process of photosynthesis where plants make their own food with carbon dioxide, using energy from the sunlight. Therefore, deforestation (cutting down of trees) has an effect on the carbon cycle also the greenhouse gas effect and global warming. Deforestation reduces the removal component of this cycle, thus, increasing the carbon dioxide in the air.

7 0

3 years ago

What is the main purpose of patent attorneys?

Aneli [31]

To protect the patents of those they work . Patents are legal rights of ownership to something that you have made or created.

Please vote my answer branliest! Thanks .

3 0

3 years ago

Read 2 more answers

How many grams of F2 gas are there in a 5.00-L cylinder at 4.00 × 10^3 mm Hg and 23°C?

GuDViN [60]

Answer:

41.17g

Explanation:

We are given the following parameters for Flourine gas(F2).

Volume = 5.00L

Pressure = 4.00× 10³mmHG

Temperature =23°c

The formula we would be applying is Ideal gas law

PV = nRT

Step 1

We find the number of moles of Flourine gas present.

T = 23°C

Converting to Kelvin

= °C + 273k

= 23°C + 273k

= 296k

V = Volume = 5.00L

R = 0.08206L.atm/mol.K

P = Pressure (in atm)

In the question, the pressure is given as 4.00 × 10³mmHg

Converting to atm(atmosphere)

1 mmHg = 0.00131579atm

4.00 × 10³ =

Cross Multiply

4.00 × 10³ × 0.00131579atm

= 5.263159 atm

The formula for number of moles =

n = PV/RT

n = 5.263159 atm × 5.00L/0.08206L.atm/mol.K × 296K

n = 1.0834112811moles

Step 2

We calculate the mass of Flourine gas

The molar mass of Flourine gas =

F2 = 19 × 2

= 38 g/mol

Mass of Flourine gas = Molar mass of Flourine gas × No of moles

Mass = 38g/mol × 1.0834112811moles

41.169628682grams

Approximately = 41.17 grams.

3 0

3 years ago

What is atomic?................................................

Pavel [41]

Answer:

oh I'm here thanks for your points

3 0

3 years ago

Read 2 more answers

a 2.7 L of N2 is collected at 121kpa and 288 K . if the pressure increases to 202 kpa and the temperature rises to 303 K , what

jok3333 [9.3K]

Answer:

The gas will occupy a volume of 1.702 liters.

Explanation:

Let suppose that the gas behaves ideally. The equation of state for ideal gas is:

$P\cdot V = n\cdot R_{u}\cdot T$ (1)

Where:

$P$ - Pressure, measured in kilopascals.

$V$ - Volume, measured in liters.

$n$ - Molar quantity, measured in moles.

$T$ - Temperature, measured in Kelvin.

$R_{u}$ - Ideal gas constant, measured in kilopascal-liters per mole-Kelvin.

We can simplify the equation by constructing the following relationship:

$\frac{P_{1}\cdot V_{1}}{T_{1}} = \frac{P_{2}\cdot V_{2}}{T_{2}}$ (2)

Where:

$P_{1}$ , $P_{2}$ - Initial and final pressure, measured in kilopascals.

$V_{1}$ , $V_{2}$ - Initial and final volume, measured in liters.

$T_{1}$ , $T_{2}$ - Initial and final temperature, measured in Kelvin.

If we know that $P_{1} = 121\,kPa$ , $P_{2} = 202\,kPa$ , $V_{1} = 2.7\,L$ , $T_{1} = 288\,K$ and $T_{2} = 303\,K$ , the final volume of the gas is:

$V_{2} = \left(\frac{T_{2}}{T_{1}} \right)\cdot \left(\frac{P_{1}}{P_{2}} \right)\cdot V_{1}$

$V_{2} = 1.702\,L$

The gas will occupy a volume of 1.702 liters.

6 0

3 years ago