Question

If the binding energy of the electron in a hydrogen atom is 13.6 eV, the energy required to remove the electron from the first excited state of $Li^{+ +}$ is

• A. 122.4 Ev
• B. 30.6 eV
• C. 13.6 Ev
• D. 3.4 Ev

Answer 1

Answer: (B)

30.6 eV

Answer 2

Using $E_{n} = - \frac{13.6 \times Z^{2}}{n^{2}}eV$

For first excited state n = 2 and for $Li^{+ +}$, Z = 3

∴ $E = - \frac{13.6}{2^{2}} \times 3^{2} = - \frac{13.6 \times 9}{4} = - 30.6eV$.

Hence, remove the electron from the first excited state of $Li^{+ +}$ be 30.6 Ev