Question

If for Li$^{2+}$ ion, electron is in transition between energy levels such that sum of principal quantum numbers is 4 and difference is 2, then find the wavelength (in cm) emitted for transition between these energy levels.
[Given: $R = 1.1 \times 10^5\ \text{cm}^{-1}$]

• A. $114 \times 10^{-8}$ cm
• B. $1026 \times 10^{-8}$ cm
• C. $12.66 \times 10^{-8}$ cm
• D. $10^{-8}$ cm

Hint:

For hydrogen-like ions, wavelength is inversely proportional to $Z^2$.

Answer 1

The Correct Option is A

Step 1: Determine the principal quantum numbers

Given:

n₁ + n₂ = 4
n₁ − n₂ = 2

Adding the two equations:

2n₁ = 6

n₁ = 3

Substituting n₁ = 3:

n₂ = 4 − 3 = 1

Thus, the electronic transition is from n₁ = 3 to n₂ = 1.

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Step 2: Use Rydberg formula for hydrogen-like ions

For a hydrogen-like ion:

1/λ = RZ² (1/n₂² − 1/n₁²)

Where:
R = 1.1 × 10⁵ cm⁻¹
Z = 3 (for Li²⁺)
n₁ = 3, n₂ = 1

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Step 3: Substitute the values

1/λ = 1.1 × 10⁵ × 3² (1/1² − 1/3²)

= 1.1 × 10⁵ × 9 × (1 − 1/9)

= 1.1 × 10⁵ × 9 × 8/9

= 1.1 × 10⁵ × 8

= 8.8 × 10⁵ cm⁻¹

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Step 4: Calculate the wavelength

λ = 1 / (8.8 × 10⁵)

λ = 114 × 10⁻⁸ cm

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Final Answer:

The wavelength of the emitted photon is
114 × 10⁻⁸ cm