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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 = 10 .

Determine the radius and energy of a He+\text{He}^+ ion in the state n=10n = 10.

Options

  1. A13.25 A˚13.25 \text{ \AA}, −1.088 eV-1.088 \text{ eV}
  2. B26.5 A˚26.5 \text{ \AA}, −0.136 eV-0.136 \text{ eV}
  3. C26.5 A˚26.5 \text{ \AA}, −0.544 eV-0.544 \text{ eV}
  4. D52.9 A˚52.9 \text{ \AA}, −0.136 eV-0.136 \text{ eV}

Correct answer

C. 26.5 A˚26.5 \text{ \AA}, −0.544 eV-0.544 \text{ eV}

Step-by-step solution

The radius of a hydrogen-like ion in the nthn^{\text{th}} state 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=10n = 10:

r10=0.529×1022=0.529×50=26.45 A˚≈26.5 A˚r_{10} = 0.529 \times \dfrac{10^2}{2} = 0.529 \times 50 = 26.45 \text{ \AA} \approx 26.5 \text{ \AA}

The energy of a hydrogen-like ion in the nthn^{\text{th}} state 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=10n = 10:

E10=−13.6×22102=−13.6×4100=−0.544 eVE_{10} = -13.6 \times \dfrac{2^2}{10^2} = -13.6 \times \dfrac{4}{100} = -0.544 \text{ eV}
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