Question

Consider two hydrogen like ions $\text{A}$ and $\text{B}.$ Ionization energy of $\text{A}$ is greater than that of $\text{B}.$ 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. $E_A>E_B$
• D. $L_A>L_B$

Answer 1

The correct option is B $u_A>u_B$

The ionization energy of a hydrogen like ion is,

$\Delta E_n=13.6\times \frac{Z^2}{n^2}\text{eV}$

Given, $\Delta E_A>\Delta E_B\Rightarrow Z_A>Z_B.......\left(1\right)$

Angular momentum in the $n^{t}h$ orbit is given by,

$L=\frac{nh}{2\pi }$

$\therefore L_A=L_B[\because n=1]$

The energy of a hydrogen like atom is,

$E=$ energy of the orbit $E\propto -\frac{Z^2}{n^2}$

From (1) we get,

$E_A<E_B$

The orbital velocity of revolving electron in the $n^{th}$ orbit is,

$u_n\propto \frac{Z}{n}$

$\therefore u_A>u_B$

The bohr radius of the $n^{th}$ orbit is,

$r_n\propto \frac{n^2}{Z}$

$\therefore r_A<r_B$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.