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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 = 4 .

Determine the radius and energy of a He+\text{He}^+ ion in the state n=4n = 4.

Options

  1. A4.24 A˚4.24 \text{ \AA}, −3.4 eV-3.4 \text{ eV}
  2. B2.12 A˚2.12 \text{ \AA}, −6.8 eV-6.8 \text{ eV}
  3. C8.46 A˚8.46 \text{ \AA}, −0.85 eV-0.85 \text{ eV}
  4. D4.24 A˚4.24 \text{ \AA}, −0.85 eV-0.85 \text{ eV}

Correct answer

A. 4.24 A˚4.24 \text{ \AA}, −3.4 eV-3.4 \text{ eV}

Step-by-step solution

The radius of the nn-th 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}^+, Z=2Z = 2 and n=4n = 4.

r4=0.529×422=0.529×8=4.232 A˚≈4.24 A˚r_4 = 0.529 \times \dfrac{4^2}{2} = 0.529 \times 8 = 4.232 \text{ \AA} \approx 4.24 \text{ \AA}

The energy of the nn-th orbit is given by

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

E4=−13.6×2242=−13.6×416=−3.4 eVE_4 = -13.6 \times \dfrac{2^2}{4^2} = -13.6 \times \dfrac{4}{16} = -3.4 \text{ eV}
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