Question

An electron makes transition from higher energy orbit to lower energy orbit in $Li^{2+}$ ion such that $n_1 + n_2 = 4$ and $n_2 - n_1 = 2$. Determine the wavelength of emitted photon in transition (in cm).

• A. $1.14 \times 10^{-6}$ cm
• B. $3.28 \times 10^{-6}$ cm
• C. $5.76 \times 10^{-6}$ cm
• D. $8.23 \times 10^{-6}$ cm

Hint:

For hydrogen-like ions, energy levels depend on $Z^2$. Always include nuclear charge.

Answer 1

The Correct Option is A

Step 1: Determine the principal quantum numbers

Given:

n₁ + n₂ = 4
n₂ − n₁ = 2

Adding the two equations:

2n₂ = 6 ⇒ n₂ = 3

Substituting in the first equation:

n₁ + 3 = 4 ⇒ n₁ = 1

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

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Step 2: Use the Rydberg formula for hydrogen-like ions

For a hydrogen-like ion:

1/λ = RZ² ( 1/n₁² − 1/n₂² )

Where:

R = 1.097 × 10⁷ m⁻¹,
Z = 3 (for Li²⁺)

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Step 3: Substitute the values

1/λ = 1.097 × 10⁷ × 3² × (1 − 1/9)

1/λ = 1.097 × 10⁷ × 9 × 8/9

1/λ = 8.776 × 10⁷ m⁻¹

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Step 4: Calculate the wavelength

λ = 1 / (8.776 × 10⁷)

λ = 1.14 × 10⁻⁸ m

Converting to centimetres:

λ = 1.14 × 10⁻⁶ cm

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Final Answer:

The wavelength of the emitted photon is
1.14 × 10⁻⁶ cm