Solveeit Logo
Login

Question

Question: A hydrogen-like atom in ground state absorbs n protons having the same energy and it emits exactly n...

A hydrogen-like atom in ground state absorbs n protons having the same energy and it emits exactly n photons when electrons transition takes place. Then, the energy of the absorbed photon may be :
(This question has multiple correct options)
(A) 91.8 eV
(B) 40.8 eV
(C) 48.4 eV
(D) 54.4 eV

Explanation

Solution

The energy related with the absorption or emission of photon can be given by following equation
ΔE=RH⋅Z2(1ni2−1nf2)J\Delta E = {R_H} \cdot {Z^2}\left( {\dfrac{1}{{{n_i}^2}} - \dfrac{1}{{{n_f}^2}}} \right)J
Where Z is the atomic number and RH{R_H} is Rydberg constant

Complete step by step solution:
We are given that the atom is like hydrogen and we will find the possible energy of the absorbed photon.
- Bohr gave the energy related with the absorption and emission of the photon in the atoms which has electronic configuration like hydrogen. So, the energy related with the hydrogen like species can be given by
ΔE=RH⋅Z2(1ni2−1nf2)J\Delta E = {R_H} \cdot {Z^2}\left( {\dfrac{1}{{{n_i}^2}} - \dfrac{1}{{{n_f}^2}}} \right)J
where Z is the atomic number and n is the principal quantum number and RH{R_H} is Rydberg constant which has the value 2.18×10−182.18 \times {10^{ - 18}} .
So, we can rewrite the above equation as
ΔE=2.18×10−18⋅Z2(1ni2−1nf2)J \Delta E = 2.18 \times {10^{ - 18}} \cdot {Z^2}\left( {\dfrac{1}{{{n_i}^2}} - \dfrac{1}{{{n_f}^2}}} \right)J{\text{ }}
Now, as the hydrogen like species absorbs photons and also emits n photons, the transition must have taken place from 1 to 2. So, putting that value in equation (1) will give
ΔE=2.18×10−18⋅Z2(112−122)J \Delta E = 2.18 \times {10^{ - 18}} \cdot {Z^2}\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}} \right)J{\text{ }}
So,
ΔE=2.18×10−18⋅Z2(34)J \Delta E = 2.18 \times {10^{ - 18}} \cdot {Z^2}\left( {\dfrac{3}{4}} \right)J{\text{ }}
Thus, we can write that
ΔE=16.35×10−19⋅Z2J\Delta E = 16.35 \times {10^{ - 19}} \cdot {Z^2}J
Now, we are given the energies in the eV unit. So, let’s convert the unit of energy from J to eV.
We know that 1.6×10−191.6 \times {10^{ - 19}}J= 1 eV
So, 16.35×10−19⋅Z2J \times {10^{ - 19}} \cdot {Z^2}J=16.35×10−191.6×10−19=10.21⋅Z2\dfrac{{16.35 \times {{10}^{ - 19}}}}{{1.6 \times {{10}^{ - 19}}}} = 10.21 \cdot {Z^2} eV
So, we obtained that the energy is 10.21⋅Z210.21 \cdot {Z^2} eV.
Now, Z is the atomic number. So, for the hydrogen-like species, the atomic number may be 2,3,…etc.
- If we put 2, then ΔE=10.21⋅Z2=10.21(2)2=40.8eV\Delta E = 10.21 \cdot {Z^2} = 10.21{(2)^2} = 40.8eV
- If we put 3, then ΔE=10.21(Z)2=10.21(3)2=91.8eV\Delta E = 10.21{(Z)^2} = 10.21{(3)^2} = 91.8eV

Hence, we can say that both options (A) and (B) are correct.

Note: Note that the ions like He+,Li2+H{e^ + },L{i^{2 + }} have electronic configuration same as hydrogen and hence they can be called hydrogen-like species. So, in the case of He+H{e^ + }, Z is 2 and in case of Li2+L{i^{2 + }}, Z is 3.