Atomic Spectra

Step 1: Energy corresponds to excitation up to n=4.
Step 2: Number of spectral lines: N=(n(n-1))/(2)=(4×3)/(2)=6

Step 1: For hydrogen-like species:
Step 2: For He , , Balmer series .
Step 3:

Step 1: Rydberg formula:
Step 2: For hydrogen Balmer line:
Step 3: For ( ):
Step 4: Solving:

Step 1: Lyman series limit corresponds to transition n=∞ → 1.
Step 2: Balmer series limit corresponds to transition n=∞ → 2.
Step 3: Since wavelength λ ∝ n²: lambdaBₐlmₑᵣ = 4lambdaLyman = 912×4AA

Step 1: First Balmer line corresponds to transition n=3 → 2.
Step 2: Rydberg formula: ν=R((1)/(2²)-(1)/(3²)) =R((1)/(4)-(1)/(9))=(5R)/(36)

Step 1: Rydberg formula: (1)/(λ) = RZ²((1)/(n₁²)-(1)/(n₂²))
Step 2: For hydrogen Balmer 2nd line: (1)/(λ) = R((1)/(2²)-(1)/(4²))
Step 3: For Li²+: R(3²)((1)/(x²)-(1)/(12²))
Step 4: Equating gives x = 5.

Step 1: Balmer series corresponds to transitions ending at n₁=2.
Step 2: First emission line is for transition: n₂=3 → n₁=2
Step 3: Using Rydberg formula: ν = R((1)/(2²)-(1)/(3²)) = R((1)/(4)-(1)/(9))
Step 4: ν = R((5)/(36))

Step 1: Rydberg formula: (1)/(λ) = R((1)/(n₁²)-(1)/(n₂²))
Step 2: Series limit corresponds to n₂→∞. For Lyman series (n₁=1): lambdaL = (1)/(R) = 912AA
Step 3: For Balmer series (n₁=2): lambdaB = (1)/(R((1)/(4))) = 4lambdaL
Step 4: lambdaB = 4×912 = 3648AA