Question

Energy of the state $\text{ }{{\text{S}}_{\text{1 }}}$in units of the hydrogen atom ground state energy is:
A) $\text{ 0}\text{.75 }$
B) $\text{ 1}\text{.50 }$
C) $\text{ 2}\text{.25 }$
D) $\text{ 4}\text{.50}$

Answer 1

Hint: The energy of a hydrogenic atom $\text{ }{{\text{E}}_{\text{n}}}\text{ }$is given as follows,
$\text{ }{{\text{E}}_{\text{n}}}\text{ = }\dfrac{K\times {{Z}^{2}}}{{{n}^{2}}}\text{ }$
Where ‘n’ is the principal quantum number and ‘Z’ is the atomic number of species and K is a Rydberg constant. The ‘n’ value can be calculated by considering the following formula,
$\text{ No}\text{.of radial nodes = }n-l-1\text{ }$
Where n is the number of total nodes, ‘l’ is the angular nodes.

Complete answer:
The hydrogen-like species for example $\text{ L}{{\text{i}}^{\text{2+}}}\text{ }$have symmetrically spherical orbits. These species have one radial node.
The quantum number ‘n’ equals the shell number of the energy level of an orbital. The energy of an energy state is related to the principal quantum number as follows,
$\text{ }{{\text{E}}_{\text{n}}}\text{ = }\dfrac{K\times {{Z}^{2}}}{{{n}^{2}}}\text{ }$ (1)
Where \[{{\text{E}}_{\text{n}}}\]is the energy of an orbital
Z is the atomic number of an atom or ion
‘n’ is the nodes or principal quantum number
Here, we are interested to find out the energy of $\text{ }{{\text{S}}_{\text{1 }}}$the energy state.
The $\text{ }{{\text{S}}_{\text{1 }}}$is a spherical and symmetrical energy state.it is spherical thus corresponds to s-orbital. Let's calculate the value of the principal quantum number for a $\text{ }{{\text{S}}_{\text{1 }}}$state. Since it is spherical it has zero angular nodes. It has one radial node .then we have,
$\begin{align}
  & \text{ Number of radial nodes = }n-l-1 \\
 & \Rightarrow 1=\text{ }n-0-1\text{ }\because l=0\text{ } \\
 & \therefore \text{ }n\text{ = 2 } \\
\end{align}$
We are interested to find out the energy of an $\text{ }{{\text{S}}_{\text{1 }}}$energy state. The $\text{ }{{\text{S}}_{\text{1 }}}$state corresponds to the
$\text{ L}{{\text{i}}^{\text{2+}}}\text{ }$ Species. The atomic number $\text{ Z= 3 }$.
 Now substitute the values in the equation (1) we have,
$\Rightarrow \text{ }{{\text{E}}_{{{\text{S}}_{1}}}}\text{ = }\dfrac{\text{K}\times {{Z}^{2}}}{{{n}^{2}}}\text{ = }\dfrac{\text{K}\times {{(3)}^{2}}}{{{(2)}^{2}}}\text{ = }\dfrac{9}{4}\text{K = 2}\text{.25 K }$
Here,$\text{ K }=\text{ }\dfrac{\text{k}{{\text{Z}}^{\text{2}}}}{{{\text{n}}^{\text{2}}}}\text{ }\left( Z=1\text{ , }n=1 \right)\text{ }$
We want the energy of a $\text{ }{{\text{S}}_{\text{1 }}}$state in the units of the hydrogen atom. Thus we have considered the energy of the hydrogen atom as K.
So, the energy of $\text{ }{{\text{S}}_{\text{1 }}}$the energy state in the units of the hydrogen atom is equal to $\text{ 2}\text{.25 K }$ or it is $\text{ 2}\text{.25 }$ times the energy of the hydrogen atom.

Hence, (C) is the correct option.

Note:
The ground state energy of a hydrogen atom is$\text{ }-13.6\text{ eV }$. If we want to measure the energy of $\text{ }{{\text{S}}_{\text{1 }}}$state it would be,
$\text{ }{{\text{E}}_{{{\text{S}}_{\text{1}}}}}\text{ = 2}\text{.25 K }=\text{ }-13.6\left( 2.25 \right)\text{ = }-30.6\text{ eV }$
The radial node is a region in which the probability of finding an electron is zero. These are regions that depict the gap between two energy levels in which electrons are allowed to transit.