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QJEEAdv16Ph3798: JEE Advanced 2016 Physics - Class 12 JEEAdv16Ph3798 Solution: Highly excited states for hydrogen-like atoms (also called…

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?

Multiple Choice Options

  • A Relative change in the radii of two consecutive orbitals does not depend on Z.
  • B Relative change in the radii of two consecutive orbitals varies as 1/n
  • C Relative change in the energy of two consecutive orbitals varies as 1/n3
  • D Relative change in the angular momenta of two consecutive orbitals varies as 1/n

Step-by-Step Verified Solution

The radius of nth n^{\text {th }} orbital for a hydrogen-like atom of atomic number ZZ is given by rn=n2a0Zr_n={{{n^2}{a_0}} \over Z} , where a0=0.53a_0=0.53 Ao\mathop A\limits^o is the Bohr's radius.

∴\therefore  Radius of orbit, r∝n2Z\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

En=−13.6Z2n2eV{E_n} = {{ - 13.6{Z^2}} \over {{n^2}}}eV

∴\therefore En∝−Z2n2E_n \propto \frac{-Z^2}{n^2}

The relative change in the energy of two consecutive orbitals is

amp;En+1−EnEn=−Z2(n+1)2+Z2n2−Z2n2=−(2n+1)(n+1)2amp;≈−2nn2=−2n(ngt;gt;1)\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 nth n^{\text {th }} orbital is given by Ln=nh2πL_n={{nh} \over {2\pi }} . The relative change in the angular momentum of two consecutive orbitals is

Ln+1−LnLn=(n+1)h2π−nh2πnh2π=1n\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}

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