Question

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. 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?

Answer 1

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