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miss Akunina [59]
3 years ago
7

According to the Bohr model, the energy of the hydrogen atom is given by the equation: E = (-21.7 x 10 -19 J)/ n 2 Calculate the

wavelength of the photon emitted when the atom undergoes relaxation from the first excited state to the ground state.
Chemistry
1 answer:
Anit [1.1K]3 years ago
7 0

Answer:

91.6 nm

Explanation:

The energy of the hydrogen atom can be calculated by the emission of a photon. When an electron is excited it goes from to the next energetic level, and when it returns to its ground state, it emits a photon. Hydrogen has only one electron, which is at the level n = 1. So, the equation is given:

E = (-21.7x10⁻¹⁹J)/1²

E = -21.7x10⁻¹⁹J

The energy of the photon is the energy absorbed, and because of that is positive (the opposite of the energy released by the electron). This energy can be calculated by:

E = h*c/λ

Where h is the Planck's constant (6.626x10⁻³⁴ J.s), c is the speed of the light (3.00x10⁸ m/s), and λ is the wavelength of the photon.

21.7x10⁻¹⁹ = 6.626x10⁻³⁴ * 3.00x10⁸/λ

λ = 9.16x10⁻⁸ m

λ = 91.6 nm

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marishachu [46]
Combustion reaction for menthol is as follows;
CxHyOz + O₂ ---> xCO₂ +  H₂O
Mass of CO₂ formed - 28.16 mg
Therefore number of moles formed - 28.16/ 44 g/mol = 0.64 mmol
Mass of water formed  - 11.53 mg
number of water moles formed - 11.53 mg/18 g/mol = 0.64 mmol
From CO₂,
1 mol of CO₂ - 1 mol of C and 2 mol of O
therefore number of C moles - 0.64 mmol
                                 O moles - 1.28 mmol
from H₂O
1 mol of H₂O - 2 mol of H and 1 mol of O
number of H moles - 1.28 mmol
                 O moles - 0.64 mmol

Mass of menthol initially - 10 mg
in reactions, the masses of products are equal to the masses of reactants. The excess mass to the products formed is due to O₂ in air
Original mass of menthol - 10 mg
mass of water and CO₂ - 11.53 mg + 28.16 mg = 39.69
Difference in mass - 39.69 - 10 = 29.69 mg
This difference comes from O moles in air - 29.69 mg/ 16 g/mol = 1.8556 mmol
then O moles coming from menthol - (1.28 + 0.64) - 1.8556 = 0.064 mmol

In menthol 
C moles - 0.64 mmol
H moles - 1.28 mmol
O moles - 0.064 mmol
ratios of C:H:O
C          H         O
0.64    1.28      0.064
x1000  x1000    x1000 to get whole numbers
640       1280      64
10          20          1
Simplest ratio of C:H:O is 10:20:1
therefore empirical formula of menthol is C₁₀H₂₀O


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