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Question: Consider a hydrogen-like ionized atom with atomic number Z with a single electron. In the emission s...

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 ____________.

Explanation

Solution

Energy of the electron in the nth orbit, En=−me48ε02ch3⋅ch⋅Z2n2=−13.6Z2n2eV{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= ε0\varepsilon _0^{}
Atomic number= Z
Transition energy for photon emitted in the mth shell to nth shell,Em−n=−13.6Z2m2−(−13.6Z2n2)eV=13.6Z2(1n2−1m2)eV{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,
E2−1=E2−E1 =−13.6Z222−(−13.6Z212)eV =13.6Z2(1−14)eV =10.2Z2eV  {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,
E3−2=E3−E2 =−13.6Z232−(−13.6Z222)eV =13.6Z2(14−19)eV =179Z2eV  {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, E2−1=E3−2+74.8{E_{2 - 1}} = {E_{3 - 2}} + 74.8
10.2Z2=179Z2+74.810.2{Z^2} = \dfrac{{17}}{9}{Z^2} + 74.8
  ⟹  \implies (10.2−179)Z2=74.8(10.2 - \dfrac{{17}}{9}){Z^2} = 74.8
  ⟹  \implies 91.8−179Z2=74.8\dfrac{{91.8 - 17}}{9}{Z^2} = 74.8
  ⟹  \implies 74.89Z2=74.8\dfrac{{74.8}}{9}{Z^2} = 74.8
  ⟹  \implies Z2=9{Z^2} = 9
∴Z=3\therefore Z = 3 as atomic number cannot be negative.
So, the value of Z is 3.

Note:
The term me48ε02ch3\dfrac{{m{e^4}}}{{8\varepsilon _0^2c{h^3}}} is known as Rydberg Constant denoted by R.
Then En=−me48ε02ch3⋅ch⋅Z2n2=−RchZ2n2{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}}}.
RchRch is the ionization energy of the hydrogen atom.
Rch=(1.09737×107)×(3×108)×(6.626×10−34)1.6×10−19=13.6eVRch = \dfrac{{(1.09737 \times {{10}^7}) \times (3 \times {{10}^8}) \times (6.626 \times {{10}^{ - 34}})}}{{1.6 \times {{10}^{ - 19}}}} = 13.6eV.