Question

The state $\text{ }{{\text{S}}_{\text{1 }}}$ is:
A) $\text{ 1s }$
B) $\text{ 2s }$
C) $\text{ 2p }$
D) $\text{ 3s }$

Answer 1

The hydrogen-like species may be spherically symmetrical. The ground state energy for a hydrogen atom $\text{ }{{\text{E}}_{\text{H}}}\text{ }$ is considered as the ${{\text{S}}_{\text{0}}}$ state. The energy for hydrogen is given as,
$\text{ }{{\text{E}}_{\text{H}}}\text{ = }\dfrac{-13.6\times {{Z}^{2}}}{{{n}^{2}}}\text{ = }-13.6\text{ }$
For the hydrogen atom, the atomic number Z is 1 and the principal quantum number n is also 1. There, n is a total number of nodes or the principal quantum number of a shell. For higher states like$\text{ }{{\text{S}}_{\text{1 }}}$,$\text{ }{{\text{S}}_{\text{2}}}\text{ }$, etc. The number of nodes ‘n’ can be calculated by considering the energy of state equal to the ground state of hydrogen atom i.e.$\text{ }{{\text{S}}_{\text{0}}}\text{ }$ state.

Complete answer:
The hydrogen-like species for example $\text{ L}{{\text{i}}^{\text{2+}}}\text{ }$have symmetrically spherical orbits. This species has one radial node. Node is defined as the point in the space at which the probability of finding the electrons has become zero.
On absorption of light radiation, this species or ions undergo the transition from the lower energy level let $\text{ }{{\text{S}}_{\text{0 }}}$ to the higher energy level$\text{ }{{\text{S}}_{\text{1 }}}$.
The $\text{ }{{\text{S}}_{\text{1 }}}$energy state also has one radial node. Thus, if we consider that $\text{ }{{\text{S}}_{\text{0 }}}$corresponds to the ground state energy level of hydrogen.
We are interested to find out the principal quantum number and the angular nodes of an $\text{ }{{\text{S}}_{\text{1 }}}$energy state. This can be calculated from the energy equation of the hydrogen atom.
Here, we will assume that the $\text{ }{{\text{S}}_{1}}\text{ }$have similar energy as that of the ground state energy of a hydrogen atom. The ground state energy for a hydrogen atom is given as,
$\text{ }{{\text{E}}_{{{\text{S}}_{1}}}}\text{ = }\dfrac{-13.6\times {{Z}^{2}}}{{{n}^{2}}}\text{ }$
Where n is the number of nodes and Z is the atomic number.
For $\text{ }{{\text{S}}_{1}}\text{ }$state, the equation is,
$\text{ }{{\text{E}}_{{{\text{S}}_{1}}}}\text{ = }\dfrac{-13.6\times {{Z}^{2}}}{{{n}^{2}}}\text{ = }{{\text{E}}_{\text{H}}}\text{ in ground state = }-13.6\text{ }$ …………………...(1)
Since for hydrogen $\text{ n = 1 }$ .
Now the $\text{ }{{\text{S}}_{1}}\text{ }$ state an atomic number equal to $\text{ Z = 2 }$ . Thus, the above equation (1) can be solved to determine the number of nodes ‘n’.We have,
$\begin{aligned} & \Rightarrow \text{ }{{\text{E}}_{{{\text{S}}_{\text{2}}}}}\text{ = }\dfrac{-13.6\times {{Z}^{2}}}{{{n}^{2}}}\text{ = }-13.6 \\\ & \text{ }\Rightarrow \text{ }{{n}^{2}}\text{ = }\dfrac{-13.6\times {{2}^{2}}}{-13.6}\text{ = 9} \\\ & \text{ }\therefore n\text{ = 2 } \\\ \end{aligned}$
Here we get that the principal quantum number is equal to 2.
So, the $\text{ }{{\text{S}}_{\text{1 }}}$ energy state has the principal quantum number equal to 2 and the ‘l’ or the angular nodes would be zero (since for $\text{ ns }$ the radial nodes equal to $\text{ n}-1\text{ }$ ). Thus the $\text{ }{{\text{S}}_{\text{1 }}}$ state is $\text{ 2s }$ orbital.

Hence, (B) is the correct option.

Note:
Note that, the number of the radial nodes (n) is always equal to the,
$\text{ }n-l-1\text{ }$ , where angular nodes (azimuthal quantum number) is represented by ‘l’.For a $\text{ }{{\text{S}}_{\text{2}}}\text{ }$ state, the n value is 3 and we know it has a radial node 1 then,
$\begin{aligned} & \text{ = }n-l-1\text{ } \\\ & \text{1 = 3}-l-1 \\\ & \therefore l=\text{ 1 } \\\ \end{aligned}$
Which means, the orbital has one angular node .Hence it is a ‘p’ subshell of which a total of three nodes .i.e. $\text{ 3p }$ state.