Solution

$E_n = -13.6\frac{Z^2}{n^2}$ eV For $He^+$ (Z = 2), n = 1: $E = -13.6 \times \frac{2^2}{1^2} = -13.6 \times 4 = -54.4$ eV $= -x$ For $Be^{3+}$ (Z = 4), n = 2: $E = -13.6 \times \frac{4^2}{2^2} = -13.6 \times 4 = -54.4$ eV So the energy for $Be^{3+}$ in n = 2 is $-x$ J.

Concept Energy scaling with atomic number and principal quantum number

Structure of Atom in past papers

62 questions from this chapter have appeared across 27 exam years.