Question

Ionization energy of a hydrogen-like ion A is greater than that of another hydrogen-like ion B. Let $r$, $u$, $E$ and $L$ represent the radius of the orbit, speed of the electron, energy of the atom and orbital angular momentum of the electron respectively. In ground state choose the incorrect option.

• A. $r_A > r_B$
• B. $u_A > u_B$
• C. $E_A > E_B$
• D. $L_A > L_B$

Answer 1

$r_A > r_B$

Answer 2

The ionization energy (IE) of a hydrogen-like ion in the ground state ($n=1$) is given by $IE = -E_1$, where $E_1$ is the ground state energy. The energy of an electron in the $n$-th orbit of a hydrogen-like ion with atomic number $Z$ is given by $E_n = -\frac{Z^2 E_0}{n^2}$, where $E_0$ is the ionization energy of hydrogen ($13.6 \text{ eV}$).

For the ground state ($n=1$), the energy is $E_1 = -Z^2 E_0$. The ionization energy from the ground state is $IE_1 = -E_1 = Z^2 E_0$.

Given $IE_A > IE_B$, it implies $Z_A > Z_B$.

• Radius of the orbit: $r = \frac{a_0}{Z}$. Thus, $r_A = \frac{a_0}{Z_A}$ and $r_B = \frac{a_0}{Z_B}$. Since $Z_A > Z_B$, we have $r_A < r_B$. Therefore, $r_A > r_B$ is incorrect.
• Speed of the electron: $u = Z v_0$. Thus, $u_A = Z_A v_0$ and $u_B = Z_B v_0$. Since $Z_A > Z_B$, we have $u_A > u_B$. Therefore, $u_A > u_B$ is correct.
• Energy of the atom: $E = -Z^2 E_0$. Thus, $E_A = -Z_A^2 E_0$ and $E_B = -Z_B^2 E_0$. Since $Z_A > Z_B$, we have $E_A < E_B$. Therefore, $E_A > E_B$ is incorrect.
• Orbital angular momentum: $L = \frac{h}{2\pi}$. Thus, $L_A = \frac{h}{2\pi}$ and $L_B = \frac{h}{2\pi}$. Therefore, $L_A = L_B$, and $L_A > L_B$ is incorrect.

Options 1, 3 and 4 are all incorrect.