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

• A. $r_A>r_B$
• B. $u_A>u_B$
• C. $L_A>L_B$
• D. None of these

Answer 1

The correct option is B $u_A>u_B$

If Ionization energy of A is greater than that of B, then Atomic number (Z) of A is greater than B.
because, for Hydrogen like atom $E=\frac{13.6Z^2}{n^2}eV$ , where $n=1$
Relation for speed of electron: $u=\sqrt{\frac{Z{k}_e{e}^2}{m_e{r}_n}}and{r}_n=\frac{n^2}{Z}\left(\frac{{\hslash }^2}{k_e{e}^2{m}_e}\right)\Rightarrow u\alpha \sqrt{Z}$
$\therefore u_A>u_B$

$r\propto \frac{1}{Z}$

Hence, $r_A<r_B$