Question

If the kinetic energy of an electron in the first orbit of $H$ atom is $13.6eV$. Then the total energy of an electron in the second orbit of ${He}^{+}$ is:

• A. $13.6eV$
• B. $3.4eV$
• C. $-13.6eV$
• D. $-3.4eV$

Answer 1

The correct option is C $-13.6eV$

$\begin{array}{c}\text{Given- * kinetic energy (KE) of e" in first orbit of}\\ \text{H-atom is}13.6eV.\\ \text{we know that -}\\ kE=13.6z^2/n^{2}n\to n^{\text{th}}\text{orbit}\\ z\to \text{atomic number.}\end{array}$

$\begin{array}{c}\text{we also know that -}\\ \text{(Total Energy)}=-(\text{kinetic Energy)}\\ \Rightarrow TE=-KE\\ \Rightarrow TE=-13.6\frac{z^2}{n^2}eV\\ \text{Now, for}H{e}^{+}\text{atom}(z=2)\text{in second}\text{orbit}(n=2)\\ TE=-13.6\text{eV}\frac{4}{4}=-13.6eV\end{array}$