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Lilit [14]
3 years ago
14

A series of lines involving a common level in the spectrum of atomic hydrogen lies at 656.46 nm, 486.27 nm, 434.17 nm, and 410.2

9 nm. What is the wavelength of the next line in the series? What is the ionization energy of the atom when it is in the lower state of the transitions?
Chemistry
1 answer:
alexandr402 [8]3 years ago
8 0

Answer:

1) the wavelength of the next line in the series is 397.2 nm

2) The ionization energy is  3.3996 eV

Explanation:

Step 1: Data given

A series of lines involving a common level in the spectrum of atomic hydrogen lies at 656.46 nm, 486.27 nm, 434.17 nm, and 410.29 nm

Step 2: Calculate n₂

The visible spectrum of light from hydrogen displays four wavelengths, 410 nm, 434 nm, 486 nm, and 656 nm,

1/λ = Rh *(1/n₁² - 1/n₂² )

⇒with λ = the wavelength

⇒with Rh = Rydberg constant for hydrogen, 1.09677583 * 10^7 m

⇒ with n₁ = the principal quantum number of an energy level

⇒with  n₂ = the principal quantum number of an energy level for the atomic electron transition

λ * Rh = n₁²* (n₁+1)² / (2n₁² + 1)

656.46 nm * 109677 cm = n₁²* (n₁+1)² / (2n₁² + 1)

7.20 = n₁²* (n₁+1)² / (2n₁² + 1)

n1 = 2

All those are in the visible spectrum and are called Balmer series, or Balmer lines.

n1 (the principal quantum number of an energy level) for Balmer series is: n1 = 2

Step 3: calculate he wavelength of the next line in the series?

1/λ = Rh *(1/n₁² - 1/n₂² )

 ⇒with  n₂ = the principal quantum number of an energy level for the atomic electron transition = 7

1/λ = 109677.6 / cm * (1/2² - 1/7²)

1/λ = 109677.6 / cm * (1/4 - 1/51.84)

λ = 397.2 nm

the wavelength of the next line in the series is 397.2 nm

Step 4: What is the ionization energy of the atom when it is in the lower state of the transitions?

The energy required to ionize the atom is:

n₂ → ∞

V∞ = 1/λ = 109677.6 / cm * (1/4 - 0)

V∞ = 109677.6 * 1 eV/ 8065.5 cm-1

V∞ = 27419.25 * 1 eV / 8065.5 cm-1

V∞ = 3.3996 eV

The ionization energy is  3.3996 eV

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