The energy of an electron in the ground state ( n = 1 ) for He^+ ion is -x J . Then, the energy for an electron in n = 2 state for Be^3+ ion in J is: - Chemistry MCQ NEET (UG) 2024 2026 | ExamTest.live

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The energy of an electron in the ground state (n = 1) for He^{+} ion is - x J . Then, the energy for an electron in n = 2 state for Be^{3 +} ion in J is:

Solution & Step-by-step Explanation

The energy of an electron in a hydrogen-like species is given by: E_{n} = - E_{0} \cdot \frac{Z ^{2}}{n ^{2}} For He^{+} (Z = 2) in ground state (n = 1): E_{He^{+}} = - E_{0} \cdot \frac{2 ^{2}}{1 ^{2}} = - 4 E_{0} Given E_{He^{+}} = - x, so - 4 E_{0} = - x \Rightarrow E_{0} = \frac{x}{4} .For Be^{3 +} (Z = 4) in n = 2 state: E_{Be^{3 +}} = - E_{0} \cdot \frac{4 ^{2}}{2 ^{2}} = - E_{0} \cdot \frac{16}{4} = - 4 E_{0} Substituting E_{0} = \frac{x}{4} : E_{Be^{3 +}} = - 4 \cdot (\frac{x}{4}) = - x J

Try it yourself before checking the explanation above.

The energy of an electron in the ground state (n = 1) for He^{+} ion is - x J . Then, the energy for an electron in n = 2 state for Be^{3 +} ion in J is:

- \frac{x}{9}

- 4 x

- \frac{4 x}{9}

- x

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