Q1 / 160 · TS EAMCET 2024 (09 May Shift 1) · Chemistry

Identify the pair of species having same energy from the following (The number given in the bracket corresponds to principal quantum number ( $n$ ) in which the electron is present)

  • A
    $\mathrm{H}(n=1), \mathrm{Li}^{2+}(n=1)$
  • Correct answer
    B
    $\mathrm{Li}^{2+}(n=3), \mathrm{Be}^{3+}(n=4)$
  • C
    $\mathrm{He}^{+}(n=1), \mathrm{Li}^{2+}(n=3)$
  • D
    $\mathrm{H}(n=3), \mathrm{Li}^{2+}(n=2)$

Solution

For a single - electron species: $\mathrm{E}_{\mathrm{n}}=-13.6 \frac{\mathrm{Z}^2}{\mathrm{n}^2} \mathrm{eV}$ Thus, $\mathrm{For}_{\mathrm{Li}^{2+}}, \mathrm{E}_3=-13.6\left(\frac{3^2}{3^2}\right)=-13.6 \mathrm{eV}$ for $\mathrm{Be}^{3+}, \mathrm{E}_4=-13.6\left(\frac{4^2}{4^2}\right)=-13.6 \mathrm{eV}$

Hence, the correct answer is option: 'B'.

Relevant Questions