Which transition of the the hydrogen spectrum would have the same length in the balmer transition n=4 to n=2 of He+ spectrum? please help me urgently.
Answers
We know that wavenumber of any radiation of any atom is given by :
= R (1/(n₁)² - 1/(n₂)²) × Z²
Given that He⁺ undergoes a balmer transition from n₂ = 4 to n₁ = 2 and Z = 2 for helium
⇒ the wavenumber of the balmer transition in He⁺ = R(1/2² - 1/4²) × 2²
⇒ the wavenumber of the balmer transition in He⁺ = R(1 - 1/4) = 3R/4
we need to find for which transition in hydrogen spectrum we get the wavenumber of the radiation as 3R/4 because it has the same length as the balmer transition n₂ = 4 to n₁ = 2 of He⁺ spectrum.
we know that atomic number of hydrogen = 1
⇒ the wavenumber of any transition in hydrogen is R (1/(n₁)² - 1/(n₂)²) × 1
this should be equal to 3R/4
⇒ R (1/(n₁)² - 1/(n₂)²) = 3R/4
⇒ (1/(n₁)² - 1/(n₂)²) = 3/4
clearly we can say that if n₁ = 1 and n₂ = 2, the above equation results 3/4
that is 1/(1)² - 1/(2)² = 3/4
Hence n₁ = 1 and n₂ = 2 transition of the hydrogen spectrsum would have the same length in the balmer transition n₂ = 4 to n₁ = 2 of He⁺ spectrum.