Question

Which of the following statements regarding the energy of the stationary state is true in the following one-electron systems ?

• A. +8.72 × 10⁻¹⁸ J for first orbit of He⁺ ion
• B. +2.18 × 10⁻¹⁸ J for second orbit of He⁺ ion
• C. -2.18 × 10⁻¹⁸ J for third orbit of Li²⁺ ion
• D. -1.09 × 10⁻¹⁸ J for second orbit of H atom.

Hint:

If $Z = n$, the energy of that orbit is always exactly equal to the energy of the first orbit of Hydrogen ($-2.18 \times 10^{-18}$ J).

Answer 1

The Correct Option is C

To determine the correct statement regarding the energy of the stationary states in one-electron systems, we need to understand the formula for the energy levels of one-electron atoms such as hydrogen-like ions. The energy of an electron in the n^th orbit of a one-electron atom or ion is given by the formula:

\(E_n = -\dfrac{Z^2 \cdot 13.6 \, \text{eV}}{n^2}\)

where:

• \(E_n\) is the energy of the n^th orbit in electron volts (eV)
• \(Z\) is the atomic number of the ion
• \(n\) is the principal quantum number (orbit number)
• The energy can be converted to Joules using the conversion factor: 1 eV = \(1.602 \times 10^{-19} \, \text{J}\)

Let's evaluate each option:

• Option 1: \(+8.72 \times 10^{-18} \, \text{J}\) for the first orbit of He⁺ ion.
  • For He⁺, \(Z = 2\) and \(n = 1\).
  • Calculate energy: \(E_1 = -\dfrac{2^2 \cdot 13.6 \, \text{eV}}{1^2} = -54.4 \, \text{eV}\)
  • Convert to Joules: \(E_1 = -54.4 \times 1.602 \times 10^{-19} \, \text{J} = -8.72 \times 10^{-18} \, \text{J}\) (should be negative).
  • Thus, the given positive energy value (+8.72 × 10⁻¹⁸ J) is incorrect.
• Option 2: \(+2.18 \times 10^{-18} \, \text{J}\) for the second orbit of He⁺ ion.
  • For He⁺, \(Z = 2\) and \(n = 2\).
  • Calculate energy: \(E_2 = -\dfrac{2^2 \cdot 13.6 \, \text{eV}}{2^2} = -13.6 \, \text{eV}\)
  • Convert to Joules: \(E_2 = -13.6 \times 1.602 \times 10^{-19} \, \text{J} = -2.18 \times 10^{-18} \, \text{J}\) (should be negative).
  • Thus, the given positive energy value (+2.18 × 10⁻¹⁸ J) is incorrect.
• Option 3: \(-2.18 \times 10^{-18} \, \text{J}\) for the third orbit of Li²⁺ ion.
  • For Li²⁺, \(Z = 3\) and \(n = 3\).
  • Calculate energy: \(E_3 = -\dfrac{3^2 \cdot 13.6 \, \text{eV}}{3^2} = -13.6 \, \text{eV}\)
  • Convert to Joules: \(E_3 = -13.6 \times 1.602 \times 10^{-19} \, \text{J} = -2.18 \times 10^{-18} \, \text{J}\) (correct value).
• Option 4: \(-1.09 \times 10^{-18} \, \text{J}\) for the second orbit of H atom.
  • For H atom, \(Z = 1\) and \(n = 2\).
  • Calculate energy: \(E_2 = -\dfrac{1^2 \cdot 13.6 \, \text{eV}}{2^2} = -3.4 \, \text{eV}\)
  • Convert to Joules: \(E_2 = -3.4 \times 1.602 \times 10^{-19} \, \text{J} = -5.44 \times 10^{-19} \, \text{J}\).
  • Thus, the given energy value (-1.09 × 10⁻¹⁸ J) is incorrect.

Thus, the correct statement is that for the third orbit of Li²⁺ ion, the energy is \(-2.18 \times 10^{-18} \, \text{J}\).