Question

Calculate the maximum and minimum wavelengths for the Balmer
series of hydrogen atom. (R = 1.097x107m-1)​

Answer 1

Given:

R = 1.097 x 10⁷ m⁻¹

To find:

Calculate the maximum and minimum wavelengths for the Balmer

series of the hydrogen atom. (R = 1.097 x 10⁷ m⁻¹)​

Solution:

We know that the Balmer series starts for n₁ = 2 to n₂ = 3, 4, 5, ..., ∞

We also know that

1/λ = R[1/n₁² - 1/n₂²]

For maximum wavelengths for the Balmer  series,

n₁ = 2 and n₂ = 3

1/λ = 1.097 x 10⁷ [1/2² - 1/3²]

1/λ = 1.097 x 10⁷ [1/4 - 1/9]

1/λ = 1.097 x 10⁷ [5/36]

1/λ = 5.485 x 10⁷ / 36

λ = 36 / (5.485 x 10⁷)

λ = 6.563 × 10⁻⁷ m

λ = 6563 Å

For minimum wavelengths for the Balmer  series,

n₁ = 2 and n₂ = ∞

1/λ = 1.097 x 10⁷ [1/2² - 1/∞²]

1/λ = 1.097 x 10⁷ [1/4 - 0]

1/λ = 1.097 x 10⁷ [1/4]

1/λ = 1.097 x 10⁷ / 4

λ = 4 / (1.097 x 10⁷)

λ = 3.646 × 10⁻⁷ m

λ = 3646 Å