Question

The wavelength of the first line in the lyman series of the spectrum of hydrogen is a1 & the first line in the balmer series of the spectrum of he+ is a2. Find the value of a2/a1

Answer 1

Answer:

a1 = 121.56 nm

a2 = 164.1 nm

a2/ a1 = 1.35

Explanation:

Given

Wavelength of hydrogen (a1) = ??

Wavelength of He⁺ (a2) = ??

a2/ a1 = ??

Solution

According to Rydberg formula

1/ λ = - R ( 1/ n²₂ - 1/n²₁) Z²

where, R is a Rydberg constant and

R = 109 677 cm ⁻¹

Atomic number (Z)

for hydrogen = 1

for helium = 2

For hydrogen

Now, from the figure in attachment

n₁ = 1 and n₂ = 2

1/ λ = - 109 677  [1/ (2)² - 1/ (1)²] .  (1)²

1/ λ = - 109 677 (- 3/4) . 1

1/ λ = 82257.75 cm ⁻¹

λ = 0.00012156 cm

 λ = 1.2156 X 10⁻⁷ m

a1 = 121.56 nm

For a2

from the figure

n₁ = 2 and n₂ = 3

1/ λ = - 109 677  [1/ (3)² - 1/ (2)²] .  (2)²

1/ λ = - 109 677 (- 5/36) . 4

1/ λ = 60931.68 cm ⁻¹

λ = 0.00016411 cm

 λ =  1.641 X 10⁻⁷ m

a2 = 164.1 nm

Now for a2/ a1 = ??

By putting values

a2/ a1 = 164.1 / 121.56

a2/ a1 = 1.35