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
