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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Bohr's Model and Physics of the Atom

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Bohr's Model and Physics of the Atom

Determine the radius and energy of a He^+ ion in the state n = 1 .

Determine the radius and energy of a He+\text{He}^+ ion in the state n=1n = 1.

Options

  1. A1.058 A˚1.058 \text{ \AA}, −3.4 eV-3.4 \text{ eV}
  2. B0.265 A˚0.265 \text{ \AA}, −13.6 eV-13.6 \text{ eV}
  3. C0.529 A˚0.529 \text{ \AA}, −13.6 eV-13.6 \text{ eV}
  4. D0.265 A˚0.265 \text{ \AA}, −54.4 eV-54.4 \text{ eV}

Correct answer

D. 0.265 A˚0.265 \text{ \AA}, −54.4 eV-54.4 \text{ eV}

Step-by-step solution

The radius of the nthn^{\text{th}} Bohr orbit for a hydrogen-like species is given by

rn=0.529×n2Z A˚r_n = 0.529 \times \dfrac{n^2}{Z} \text{ \AA}

For He+\text{He}^+, the atomic number Z=2Z = 2. For the state n=1n = 1,

r1=0.529×122=0.2645 A˚≈0.265 A˚r_1 = 0.529 \times \dfrac{1^2}{2} = 0.2645 \text{ \AA} \approx 0.265 \text{ \AA}

The energy of the nthn^{\text{th}} state for a hydrogen-like species is given by

En=−13.6×Z2n2 eVE_n = -13.6 \times \dfrac{Z^2}{n^2} \text{ eV}

For He+\text{He}^+ in the state n=1n = 1,

E1=−13.6×2212=−13.6×4=−54.4 eVE_1 = -13.6 \times \dfrac{2^2}{1^2} = -13.6 \times 4 = -54.4 \text{ eV}
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