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A hydrogen-like atom with atomic number Z is in an excited state with principal quantum number 2n. The maximum energy photon it can emit is 204 eV. When the atom transitions to quantum number n, it emits a photon of energy 40.8 eV. Identify the correct statement(s).

1. The value of n is 2.
2. The value of Z is 3.
3. The minimum energy photon emitted during de-excitation of this atom is 10.57 eV.
4. The ground state energy of this atom is -217.6 eV.

Correct answer: The value of n is 2.

Solution

From level 2n to level n: E = 13.6 Z² [1/n² - 1/(2n)²] = 13.6 Z² (3)/(4n²) = 40.8 eV. From level 2n to ground (level 1): E_max = 13.6 Z² [1 - 1/(4n²)] = 204 eV. Dividing: [1 - 1/(4n²)] / [3/(4n²)] = 5, giving 4n² - 1 = 15, so n = 2. Then Z² = 40.8 x 16 / (3 x 13.6) = 16, Z = 4. Ground state energy = -13.6 x 16 = -217.6 eV. Minimum photon: from level 4 to level 3 = 13.6 x 16 x (1/9 - 1/16) = 217.6 x 7/144 = 10.57 eV.

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