Question

The hydrogen-like species $L{i^{ + 2}}$ is in a spherically symmetric state ${S_1}$ with one radial node and its energy is equal to the ground state of the hydrogen atom.
The energy of the state ${S_2}$ (next state) in units of the hydrogen atom ground state energy is:
A) 7.5
B) 1.50
C) 2.25
D) 4.50

Answer 1

Hint: The ground state of Hydrogen atom is considered as ${S_0}$ and the energy of Hydrogen atom is depicted as ${E_H} = - 13.6\dfrac{{{Z^2}}}{{{n^2}}} = - R\dfrac{{{Z^2}}}{{{n^2}}}$
In here Z is the atomic number and n is the no. of nodes. For Hydrogen atoms both Z and n will be unity itself. For finding the energies of Higher states like ${S_1}$ , ${S_2}$ , it can be found out by considering the energy of that state to be equal to the energy of the ground state ${S_0}$ .

Complete answer: We have been given the information that $L{i^{ + 2}}$ is in a spherically symmetric state. The only spherically symmetrical orbital is the s-orbital. The azimuthal quantum number l for s orbital is Zero. It has also been given that there is only a radial node present. The formula for finding the no. of radial nodes can be given as: $r = n - l - 1$ -- (1)
Where r is the no. of radial nodes, n is the principal quantum number and l is the azimuthal quantum number. Therefore, for ground state ${S_1}$ , the principal quantum number ‘n’ can be given as:
 $1 = n - 0 - 1 \to n = 2$
 If n is 2 and l is 0, we can say that the orbital is 2s. Hence ${S_1}$ is in 2s orbital.
Firstly, let us find the ground state energy of Hydrogen Atom: ${E_H} = - R\dfrac{{{1^2}}}{{{1^2}}} = - R$ -- (2) ( since the value of n and Z for Hydrogen atom is 1 )
The energy of ${S_1}$ orbital can be given as: $ - R\dfrac{{{Z^2}}}{{{n^2}}}$
Considering Z=3 for $L{i^{ + 2}}$ and also that energy of ${S_1}$ will be equal to the ground state energy of Hydrogen atom: $ - R\dfrac{{{1^2}}}{{{1^2}}} = - R\dfrac{{{3^2}}}{{{n^2}}} \to n = 3$
Note that it has only one radial node, therefore finding the value of l from the formula (1) given above; $1 = 3 - l - 1 \to l = 1$
L will be 1 for p orbital, hence the orbital ${S_2}$ will be 3p orbital.
Finding the energy of ${S_2}$ orbital $ = - R\dfrac{{{Z^2}}}{{{n^2}}} = - R\dfrac{{{3^2}}}{{{2^2}}} = - \dfrac{9}{4}R = - 2.25R$-- (3)
Comparing equation (2) with (3) we get the energy of ${S_2}$ orbital in terms of energy of Hydrogen atom.
Energy of ${S_2}$ orbital $ = 2.25{E_H}$
Hence the correct answer is Option C.

Note:
Note that the values of azimuthal quantum numbers for different orbitals have to be remembered. If l is zero then it is surely s-orbital. If l is 1 then it is p-orbital, if l is 2 then it is d orbital. If we know the number of radial nodes, we can use the formula $r = n - l - 1$ to find the azimuthal quantum number.