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Chemistry · Structure of Atom

JEE Main 2025 — 8 April, Evening Shift — Question 20

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+\mathrm{Be}^{3+} is _____\_\_\_\_\_ eV (nearest integer value)

Answer: 54

Numerical answer — enter this value.

Step-by-step solution

E1\mathrm{E}_{1} of H -atom =−13.6eV=-13.6 \mathrm{eV}

E2 of Be3+=−13.6×Z2n2=−13.6×(4)2(2)2=−54.4eV\begin{aligned} \mathrm{E}_{2} \text { of } \mathrm{Be}^{3+} & =\frac{-13.6 \times \mathrm{Z}^{2}}{\mathrm{n}^{2}} \\& =\frac{-13.6 \times(4)^{2}}{(2)^{2}} \\& =-54.4 \mathrm{eV} \end{aligned}

∣E2∣\left|E_{2}\right| of Be3+=54eV\mathrm{Be}^{3+}=54 \mathrm{eV}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
8 April, Evening Shift
Subject
Chemistry
Chapter
Structure of Atom
Topic
Bohr's Model of Atom

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