JEE Advance - Physics (2016 - Paper 1 Offline - No. 13)
Highly excited states for hydrogen-like atoms (also called Rydberg states) with nuclear charge Ze are defined by their principle quantum number n, where n >> 1. Which of the following statement(s) is(are) true?
Relative change in the radii of two consecutive orbitals does not depend on Z.
Relative change in the radii of two consecutive orbitals varies as 1/n
Relative change in the energy of two consecutive orbitals varies as 1/n3
Relative change in the angular momenta of two consecutive orbitals varies as 1/n
Giải thích
The radius of $n^{\text {th }}$ orbital for a hydrogen-like atom of atomic number $Z$ is given by $r_n={{{n^2}{a_0}} \over Z}$, where $a_0=0.53 $$$\mathop A\limits^o $$ is the Bohr's radius.
$$ \therefore $$ $$ \text { Radius of orbit, } r \propto \frac{n^2}{Z} $$
The relative change in the radii of two consecutive orbitals is
$$ \frac{r_{n+1}-r_n}{r_n}=\frac{\frac{(n+1)^2}{Z}-\frac{n^2}{Z}}{\frac{n^2}{Z}}=\frac{2 n+1}{n^2} \approx \frac{2}{n} \quad(\because n>>1) $$
The energy of the nth orbital is given by
$${E_n} = {{ - 13.6{Z^2}} \over {{n^2}}}eV$$
$$ \therefore $$ $$ E_n \propto \frac{-Z^2}{n^2} $$
The relative change in the energy of two consecutive orbitals is
$$ \begin{aligned} & \frac{E_{n+1}-E_n}{E_n}=\frac{\frac{-Z^2}{(n+1)^2}+\frac{Z^2}{n^2}}{\frac{-Z^2}{n^2}}=-\frac{(2 n+1)}{(n+1)^2} \\\\ & \approx-\frac{2 n}{n^2}=-\frac{2}{n} \quad(n>>1) \end{aligned} $$
The angular momentum of the $n^{\text {th }}$ orbital is given by $L_n={{nh} \over {2\pi }}$. The relative change in the angular momentum of two consecutive orbitals is
$$ \frac{L_{n+1}-L_n}{L_n}=\frac{(n+1) {{h} \over {2\pi }}-{{nh} \over {2\pi }}}{{nh} \over {2\pi }}=\frac{1}{n} $$
$$ \therefore $$ $$ \text { Radius of orbit, } r \propto \frac{n^2}{Z} $$
The relative change in the radii of two consecutive orbitals is
$$ \frac{r_{n+1}-r_n}{r_n}=\frac{\frac{(n+1)^2}{Z}-\frac{n^2}{Z}}{\frac{n^2}{Z}}=\frac{2 n+1}{n^2} \approx \frac{2}{n} \quad(\because n>>1) $$
The energy of the nth orbital is given by
$${E_n} = {{ - 13.6{Z^2}} \over {{n^2}}}eV$$
$$ \therefore $$ $$ E_n \propto \frac{-Z^2}{n^2} $$
The relative change in the energy of two consecutive orbitals is
$$ \begin{aligned} & \frac{E_{n+1}-E_n}{E_n}=\frac{\frac{-Z^2}{(n+1)^2}+\frac{Z^2}{n^2}}{\frac{-Z^2}{n^2}}=-\frac{(2 n+1)}{(n+1)^2} \\\\ & \approx-\frac{2 n}{n^2}=-\frac{2}{n} \quad(n>>1) \end{aligned} $$
The angular momentum of the $n^{\text {th }}$ orbital is given by $L_n={{nh} \over {2\pi }}$. The relative change in the angular momentum of two consecutive orbitals is
$$ \frac{L_{n+1}-L_n}{L_n}=\frac{(n+1) {{h} \over {2\pi }}-{{nh} \over {2\pi }}}{{nh} \over {2\pi }}=\frac{1}{n} $$