Question

Consider a hydrogen-like ionized atom with atomic number Z with a single electron. In the emission spectrum of this atom, the photon emitted in the n = 2 to n = 1 transition has energy 74.8eV higher than the photon emitted in the n = 3 to n = 2 transition. The ionization energy of the hydrogen atom is 13.6eV. The value of Z is ____________.

Answer 1

Hint: Energy of the electron in the nth orbit, ${E_n} = - \dfrac{{m{e^4}}}{{8\varepsilon _0^2c{h^3}}} \cdot ch \cdot \dfrac{{{Z^2}}}{{{n^2}}} = - \dfrac{{13.6{Z^2}}}{{{n^2}}}eV$
Where, mass of the electron= m
Charge of an electron= e
Velocity of light= c
Planck’s constant= h
Permittivity of vacuum= $\varepsilon _0^{}$
Atomic number= Z
Transition energy for photon emitted in the mth shell to nth shell,${E_{m - n}} = - \dfrac{{13.6{Z^2}}}{{{m^2}}} - ( - \dfrac{{13.6{Z^2}}}{{{n^2}}})eV = 13.6{Z^2}(\dfrac{1}{{{n^2}}} - \dfrac{1}{{{m^2}}})eV$
We have to equate the relation between transition energies of two different transitions mentioned in the problem.

Complete step by step answer:
Transition energy for photon emitted in the n= 2 to n= 1,
$
  {E_{2 - 1}} = {E_2} - {E_1} \\
   = - \dfrac{{13.6{Z^2}}}{{{2^2}}} - ( - \dfrac{{13.6{Z^2}}}{{{1^2}}})eV \\
   = 13.6{Z^2}(1 - \dfrac{1}{4})eV \\
   = 10.2{Z^2}eV \\
 $
Transition energy for photon emitted in the n= 3 to n= 2,
$
  {E_{3 - 2}} = {E_3} - {E_2} \\
   = - \dfrac{{13.6{Z^2}}}{{{3^2}}} - ( - \dfrac{{13.6{Z^2}}}{{{2^2}}})eV \\
   = 13.6{Z^2}(\dfrac{1}{4} - \dfrac{1}{9})eV \\
   = \dfrac{{17}}{9}{Z^2}eV \\
 $
According to the problem, ${E_{2 - 1}} = {E_{3 - 2}} + 74.8$
$10.2{Z^2} = \dfrac{{17}}{9}{Z^2} + 74.8$
$\implies$ $(10.2 - \dfrac{{17}}{9}){Z^2} = 74.8$
$\implies$ $\dfrac{{91.8 - 17}}{9}{Z^2} = 74.8$
$\implies$ $\dfrac{{74.8}}{9}{Z^2} = 74.8$
$\implies$ ${Z^2} = 9$
$\therefore Z = 3$ as atomic number cannot be negative.
So, the value of Z is 3.

Note:
The term $\dfrac{{m{e^4}}}{{8\varepsilon _0^2c{h^3}}}$ is known as Rydberg Constant denoted by R.
Then ${E_n} = - \dfrac{{m{e^4}}}{{8\varepsilon _0^2c{h^3}}} \cdot ch \cdot \dfrac{{{Z^2}}}{{{n^2}}} = - Rch\dfrac{{{Z^2}}}{{{n^2}}}$.
$Rch$ is the ionization energy of the hydrogen atom.
$Rch = \dfrac{{(1.09737 \times {{10}^7}) \times (3 \times {{10}^8}) \times (6.626 \times {{10}^{ - 34}})}}{{1.6 \times {{10}^{ - 19}}}} = 13.6eV$.