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zaharov [31]
2 years ago
10

A sample of excited atoms lie 3.314×10−19 J above the ground state. Determine the emission wavelength of these atoms.

Chemistry
1 answer:
BabaBlast [244]2 years ago
8 0

The emission wavelength of the atoms excited above the ground state is 599.8 nm.

<h3>Emission wavelength of the atoms</h3>

The emission wavelength of the atom is determined by using the following formulas as shown below;

E = hf

E = hc/λ

where;

  • E is the energy of the atom
  • h is Planck's constant
  • c is speed of light
  • λ is the emission wavelength

λ = hc/E

λ = (6.626 x 10⁻³⁴ x 3 x 10⁸) / (3.314 x 10⁻¹⁹)

λ = 5.998 x 10⁻⁷ m

λ = 599.8 x 10⁻⁹ m

λ = 599.8 nm

Thus, the emission wavelength of the atoms is 599.8 nm.

Learn more about wavelength here: brainly.com/question/10728818

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Explanation:

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=> Moles of Hydrogen

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Mole ratio of Sulfur to Hydrogen = 3 : 4,

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Read 2 more answers
Calculate the [OH-] and the pH of 0.035 M KOH.
DanielleElmas [232]

Answer:

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[ H_3O^+] [OH^-] = k_w

[ H_3O^+] = \frac{k_w}{ [OH^-]}

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pH =

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pH

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4 years ago
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3 0
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