The energy of an electron in first Bohr orbit of H-atom is –13.6 eV. The magnitude of energy value of electron in the first excited state of Be³⁺ is _____ eV (nearest integer value).

 Question:

The energy of an electron in the first Bohr orbit of H-atom is –13.6 eV. The magnitude of energy value of electron in the first excited state of Be³⁺ is ____ eV (nearest integer value).

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Short Solution (Text):

Step 1: Formula for energy in Bohr model

$$E_n=-13.6\frac{Z^2}{n^2}\text{eV}$$

where $Z=\; atomic\; number,$$n=\; principal\; quantum\; number.$

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Step 2: For Be³⁺ ion

• Atomic number of Be = 4.

• For Be³⁺ (hydrogen-like species), $Z = 4$.

• First excited state ⇒ $n = 2$.

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Step 3: Energy calculation

$$E_2 = -13.6 \times \frac{4^2}{2^2}$$ $$E_2 = -13.6 \times \frac{16}{4}$$ $$E_2 = -13.6 \times 4 = -54.4 \, \text{eV}$$

Magnitude = 54.4 eV

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✅ Final Answer:

$$\text{Magnitude of energy = 54 eV (nearest integer)}$$

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Conclusion – Video Solution:

Using Bohr’s energy level formula for hydrogen-like species, the energy magnitude of the electron in the first excited state (n = 2) of Be³⁺ comes out to be ≈ 54 eV.