JEE Challenger
More from Atoms
PhyMedAdv 2008 • Paper 1

Wavelength of Emitted Visible Light from Helium Ion

Comprehension Passage

Paragraph for Question Nos. 41 to 43

In a mixture of H−He+\text{H} - \text{He}^+ gas (He+\text{He}^+ is singly ionized He atom), H atoms and He+\text{He}^+ ions are excited to their respective first excited states. Subsequently, H atoms transfer their total excitation energy to He+\text{He}^+ ions (by collisions). Assume that the Bohr model of atom is exactly valid.

The wavelength of light emitted in the visible region by He+\text{He}^+ ions after collisions with H atoms is

Options

A

6.5×10−7 m6.5 \times 10^{-7} \text{ m}

B

5.6×10−7 m5.6 \times 10^{-7} \text{ m}

C

4.8×10−7 m4.8 \times 10^{-7} \text{ m}

Correct
D

4.0×10−7 m4.0 \times 10^{-7} \text{ m}

Topics & Concepts

AtomsBohr Model

Step-by-Step Solution

To find the wavelength of light emitted in the visible region by the He+\text{He}^+ ions, we proceed with the following steps:

1. Energy Transferred by the Hydrogen Atoms

The energy levels of a hydrogen atom (Z=1Z = 1) are given by:
En(H)=−13.6n2 eVE_n(\text{H}) = -\frac{13.6}{n^2} \text{ eV}

Initially, the hydrogen atoms are in their first excited state (n=2n = 2). The total excitation energy of an excited hydrogen atom is the energy required to excite it from its ground state (n=1n = 1) to n=2n = 2:
ΔEexc(H)=E2−E1=−13.622−(−13.612)=−3.4−(−13.6)=10.2 eV\Delta E_{\text{exc}}(\text{H}) = E_2 - E_1 = -\frac{13.6}{2^2} - \left(-\frac{13.6}{1^2}\right) = -3.4 - (-13.6) = 10.2 \text{ eV}

2. State of the He+\text{He}^+ Ions After Collision

The energy levels of a singly ionized helium atom He+\text{He}^+ (Z=2Z = 2) are:
En(He+)=−13.6×Z2n2=−54.4n2 eVE_n(\text{He}^+) = -13.6 \times \frac{Z^2}{n^2} = -\frac{54.4}{n^2} \text{ eV}

Initially, the He+\text{He}^+ ions are also in their first excited state (n=2n = 2):
E2(He+)=−54.422=−13.6 eVE_2(\text{He}^+) = -\frac{54.4}{2^2} = -13.6 \text{ eV}

When the hydrogen atom transfers its total excitation energy of 10.2 eV10.2\text{ eV} to a He+\text{He}^+ ion via collision, the new energy of the He+\text{He}^+ ion becomes:
Efinal(He+)=E2(He+)+ΔEexc(H)=−13.6 eV+10.2 eV=−3.4 eVE_{\text{final}}(\text{He}^+) = E_2(\text{He}^+) + \Delta E_{\text{exc}}(\text{H}) = -13.6\text{ eV} + 10.2\text{ eV} = -3.4\text{ eV}

Let the principal quantum number of this final state be nn. Then:
−54.4n2=−3.4  ⟹  n2=54.43.4=16  ⟹  n=4-\frac{54.4}{n^2} = -3.4 \implies n^2 = \frac{54.4}{3.4} = 16 \implies n = 4

Thus, the He+\text{He}^+ ions are excited to the state n=4n = 4.

3. Transitions and Identification of Visible Light Emission

From the state n=4n = 4, the possible radiative transitions for He+\text{He}^+ are:

  • 4→14 \to 1: UV region (λ≈24.3 nm\lambda \approx 24.3\text{ nm})
  • 4→24 \to 2: UV region (λ≈121.5 nm\lambda \approx 121.5\text{ nm})
  • 3→13 \to 1: UV region (λ≈25.6 nm\lambda \approx 25.6\text{ nm})
  • 3→23 \to 2: UV region (λ≈164.1 nm\lambda \approx 164.1\text{ nm})
  • 2→12 \to 1: Extreme UV region (λ≈30.4 nm\lambda \approx 30.4\text{ nm})
  • 4→34 \to 3: Visible region (wavelength lying between 380 nm380\text{ nm} and 750 nm750\text{ nm})

Thus, the only transition that emits radiation in the visible spectrum is from n=4n = 4 to n=3n = 3.

4. Calculation of the Emitted Wavelength

Using the Rydberg formula for a hydrogen-like ion with Z=2Z = 2:
1λ=RZ2(1n12−1n22)\frac{1}{\lambda} = R Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)

Substituting Z=2Z = 2, n1=3n_1 = 3, n2=4n_2 = 4, and the Rydberg constant R≈1.097×107 m−1R \approx 1.097 \times 10^7\text{ m}^{-1}:
1λ=R×22×(132−142)=4R(19−116)=4R×7144=7R36\frac{1}{\lambda} = R \times 2^2 \times \left(\frac{1}{3^2} - \frac{1}{4^2}\right) = 4R \left(\frac{1}{9} - \frac{1}{16}\right) = 4R \times \frac{7}{144} = \frac{7R}{36}

Therefore:
λ=367R=367×1.097×107 m−1≈4.69×10−7 m≈4.8×10−7 m\lambda = \frac{36}{7R} = \frac{36}{7 \times 1.097 \times 10^7 \text{ m}^{-1}} \approx 4.69 \times 10^{-7}\text{ m} \approx 4.8 \times 10^{-7}\text{ m}

Hence, the correct option is (C).

PreviousNext