Question

The Balmer series in the hydrogen spectrum corresponds to the transition from n1=2 to n2=3,4……. This series lies in the visible region. Calculate the wave number of line associated with the transition in Balmer series when the electron moves to n=4 orbit. (RH = 109677 cm-1).

Answer 1

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Answer 2

Answer- The above question is from the chapter 'Structure of Atom'.

Concept used: 1) $\frac{1}{\lambda}$ = ṽ

2) $\dfrac{1}{\lambda} = R[\dfrac{1}{n_{1}^{2}} - \dfrac{1}{n_{2}^{2}} ] \: \times Z^{2}$

Given question: The Balmer series in the Hydrogen spectrum corresponds to the transition from n₁ = 2 to n₂ = 3, 4, … . This series lies in the visible region. Calculate the wave number of line associated with the transition in Balmer series when the electron moves to n = 4 orbit. (R = 109677 cm⁻¹)

Answer: For Hydrogen atom,

Z = 1 (Atomic Number)

R = 109677 cm⁻¹ (Rydberg's Constant)

n₁ = 2

n₂ = 4

By using the formula,

$\dfrac{1}{\lambda} = R[\dfrac{1}{n_{1}^{2}} - \dfrac{1}{n_{2}^{2}} ] \: \times Z^{2}$

$\dfrac{1}{\lambda} = R[\dfrac{1}{2^{2}} - \dfrac{1}{4^{2}} ] \: \times 1^{2}$

$\dfrac{1}{\lambda} = R \: [\dfrac{4 \: - \: 1}{16}]$

$\dfrac{1}{\lambda} = R \: [\dfrac{3}{16}]$

$\dfrac{1}{\lambda} = 109677 \: \times \: \dfrac{3}{16}$

$\dfrac{1}{\lambda} = \dfrac{329031}{16}$

$\dfrac{1}{\lambda} = 20,564.4375$

We know that   .

⇒ $\bar{\nu} =  20,564.4375 \: cm^{-1}$

∴ Wave number = 20,564.4375 cm⁻¹