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A hydrogen-like atom has a ground state binding energy of 122.4 eV. Which of the following statements about this atom are correct? (A) The atomic number Z of this atom is 3. (B) The energy required to excite the atom from ground state to its first excited state is 91.8 eV. (C) The wavelength of photon emitted when the atom transitions from the first excited state to the ground state falls in the ultraviolet region. (D) The ionization energy from the first excited state is 30.6 eV.

1. (A) and (D) only
2. (A), (B), and (D) only
3. (A), (C), and (D) only
4. All of (A), (B), (C), and (D)

Correct answer: (A), (C), and (D) only

Solution

Ground state binding energy = 13.6 * Z² = 122.4 eV gives Z² = 9, so Z = 3 (Lithium-like ion, Li²+). E₁ = -122.4 eV, E₂ = -122.4/4 = -30.6 eV. Energy to go from n=1 to n=2: 122.4 - 30.6 = 91.8 eV (B is correct). Photon energy for n=2 to n=1 transition: 91.8 eV corresponds to wavelength = 1240/91.8 nm = 13.5 nm, which is in the extreme UV/soft X-ray range, so (C) is correct that it is in the UV region. Ionization from first excited state (n=2): 30.6 eV, so (D) is correct. Statement (B) about excitation energy is also 91.8 eV which is correct. All four are correct.

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