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Problems · Problem 2.11

Q.Calculate the energy associated with the first orbit of He+He^+. What is the radius of this orbit?

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For a hydrogen-like ion, the energy of the nn-th orbit is En=−13.6Z2n2E_n = -13.6 \frac{Z^2}{n^2} eV and the radius is rn=0.529n2Zr_n = 0.529 \frac{n^2}{Z} Å. For He+He^+ (Z=2Z=2) in the first orbit (n=1n=1), the energy is −54.4-54.4 eV and the radius is 0.26450.2645 Å.

The key to solving this lies in understanding energy level quantization in hydrogen-like atoms. Bohr’s model, though not the full quantum picture, gives exact results for one-electron ions like He+He^+ (helium nucleus with one electron removed). The electron is bound to a nucleus of charge +Ze+Ze, where ZZ is the atomic number. For helium, Z=2Z=2.

Why does this matter? Because the Coulomb attraction is stronger than in hydrogen (Z=1Z=1), so the electron is pulled in tighter — the orbits are smaller and the binding energy is larger. The formulas for hydrogen simply scale with Z2Z^2 for energy and 1/Z1/Z for radius.

Let’s work through it step by step.

  1. Recall the standard formulas for a hydrogen-like atom.

    For an electron in the nn-th orbit:

    • Energy: En=−13.6 Z2n2 eVE_n = - \frac{13.6 \, Z^2}{n^2} \ \text{eV}
    • Radius: rn=0.529 n2Z A˚r_n = \frac{0.529 \, n^2}{Z} \ \text{Å} These come from balancing centripetal force with Coulomb force and quantizing angular momentum. The constants 13.613.6 eV (Rydberg energy) and 0.5290.529 Å (Bohr radius) are for hydrogen (Z=1Z=1, n=1n=1).

    En=−13.6Z2n2 eV,rn=0.529n2Z A˚E_n = -13.6 \frac{Z^2}{n^2} \ \text{eV}, \quad r_n = 0.529 \frac{n^2}{Z} \ \text{Å}

  2. Identify the parameters for He+He^+.

    • Atomic number: Z=2Z = 2
    • Orbit number: n=1n = 1 (first orbit, ground state)
  3. Calculate the energy.

    Substitute Z=2Z=2, n=1n=1 into the energy formula:

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

The negative sign means the electron is bound — you’d need to supply +54.4+54.4 eV to ionize it.

Watch out

A common mistake is to forget squaring ZZ or to use Z=1Z=1 out of habit. For He+He^+, Z=2Z=2, so the energy is four times that of hydrogen’s ground state (−13.6-13.6 eV), not double. …

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