Question

A hydrogen-like atom of atomic number Z is in an excited state of quantum number $2n.$ It can emit a maximum energy photon of $204eV$. If it makes a transition to a quantum state $n,$ a photon of energy $40.5eV$ is emitted. Find n, Z and the ground state energy (in eV) for his atom. Also, calculate the minimum energy (in eV) that can be emitted by this atom during de-excitation. Ground state energy of hydrogen atom is $ - 13.6\,eV.$
A. \[{\text{3,4,10}}{\text{.5 eV}}{\text{.}}\]
B. ${\text{8,4,10}}{\text{.5 eV}}{\text{.}}$
C. ${\text{2,4,10}}{\text{.5 eV}}{\text{.}}$
D. ${\text{2,4,15}}{\text{.5 eV}}{\text{.}}$

Answer 1

Hint: The quantum of energy is released when an electron jumps from the higher energy level to the lower energy level. First find the energy at the ground state and later further in the solutions use the given conditions and solve accordingly finding the correlation between the known and unknown terms.

Complete step by step answer:
Let the energy at the ground state be $ = {E_1}$
Energy released with the given condition –
${E_{2n}} - {E_1} = 204eV$
$\dfrac{{{E_1}}}{{4{n^2}}} - {E_1} = 204eV$
Take common terms from the left hand side of the equation –
${E_1}(\dfrac{1}{{4{n^2}}} - 1) = 204eV{\text{ }}.....{\text{ (a)}}$
Also, given that it makes a transition to quantum state $n,$ a photon of energy $40.5eV$ is emitted.
$
  {E_{2n}} - {E_n} = 40.8eV \\
  \dfrac{{{E_1}}}{{4{n^2}}} - \dfrac{{{E_1}}}{{{n^2}}} = 40.8eV \\
 $
Take ${E_1}$ common from left hand side of the equation –
${E_1}\left( {\dfrac{1}{{4{n^2}}} - \dfrac{1}{{{n^2}}}} \right) = 40.8eV$
Simplify the above equation –
${E_1}\left( {\dfrac{{ - 3}}{{4{n^2}}}} \right) = 40.8eV{\text{ }}....{\text{(b)}}$
From the equations (a) and (b)
$\dfrac{{1 - \dfrac{1}{{4{n^2}}}}}{{\dfrac{3}{{4{n^2}}}}} = 5{\text{ or 1 = }}\dfrac{1}{{4{n^2}}} + \dfrac{{15}}{{4{n^2}}}$
Simplify the above equation –
$ \Rightarrow \dfrac{4}{{{n^2}}} = 1{\text{ or n = 2}}$
Equation (b) implies –
${E_1} = \dfrac{{ - 4{n^2}}}{3}40.8eV{\text{ }}$
Place $n = 2$ in the above equation –
$
  {E_1} = \dfrac{{ - 4{{(2)}^2}}}{3}40.8eV{\text{ }} \\
  {E_1} = \dfrac{{ - 4(4)}}{3}40.8eV{\text{ }} \\
  \implies {E_1}{\text{ = - 217}}{\text{.6eV }} \\
 $
Also, given that –
${E_1} = - (13.6){z^2}$
Make unknown ${z^2}$ the subject –
$
  \dfrac{{{E_1}}}{{ - (13.6)}} = {z^2} \\
  {z^2} = \dfrac{{ - {E_1}}}{{(13.6)}} \\
 $
Now place the values of ${E_1}$
${z^2} = \dfrac{{ - 217.6}}{{ - 13.6}}$
Negative sign on the numerator and the denominator cancels each other.
\[
  {z^2} = \dfrac{{217.6}}{{13.6}} \\
  {z^2} = 16 \\
 \]
Take square root on both the sides of the equation –
$z = 4$
Now, the minimum energy
$
  {E_{\min }} = {E_{2n}} - {E_{2n - 1}} \\
  {E_{\min }} = \dfrac{{{E_1}}}{{4{n^2}}} - \dfrac{{{E_1}}}{{{{(2n - 1)}^2}}} \\
 $
Place $n = 2$
$
  {E_{\min }} = \dfrac{{{E_1}}}{{4{{(2)}^2}}} - \dfrac{{{E_1}}}{{{{(2(2) - 1)}^2}}} \\
  {E_{\min }} = \dfrac{{{E_1}}}{{16}} - \dfrac{{{E_1}}}{9} \\
 $
Place the value of ${\text{ }}{E_1}{\text{ = - 217}}{\text{.6eV }}$in the above equations –
${E_{\min }} = \dfrac{{ - 217.6}}{{16}} - \dfrac{{ - 217.6}}{9}$
Minus and minus becomes plus
$
  {E_{\min }} = \dfrac{{ - 217.6}}{{16}} + \dfrac{{217.6}}{9} \\
 \implies {E_{\min }} = - 13.6 + 24.17 \\
\implies {E_{\min }} = 10.5777 \\
\implies {E_{\min }} = 10.58eV \\
 $

So, the correct answer is “Option C”.

Note:
In all the transitions, electrons emit one quantum (one photon) of the energy, but the amount of energy of each of the quantum differs whenever the electron jumps from the higher to the lower level energy is emitted.