The ground state energy of hydrogen atom is -13.6 eV . Consider an electronic...

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The ground state energy of hydrogen atom is

- 13.6 e V

$- 13.6 e V$ . Consider an electronic state

Ψ

$Ψ$ of

{H e}^{+}

${H e}^{+}$ whose energy, azimuthal quantum number and magnetic quantum number are

- 3.4 e V, 2

$- 3.4 e V, 2$ and

0

$0$ respectively. Which of the following statement(s) is(are) true for the state

Ψ

$Ψ$ ?

It has

2

$2$ angular nodes

It has

3

$3$ radial nodes

It is a

4 d

$4 d$ state

The nuclear charge experienced by the electron in this state is less than

2 e

$2 e$ , where e is the magnitude of the electronic charge.

Solution:

∵ E_{H e^{+}}^{+} = \frac{- 13.6 (z^{2})}{n^{2}} = \frac{- 13.6 {(2)}^{2}}{n^{2}} = - 3.4

$∵ E_{H e^{+}} = \frac{- 13.6 (z^{2})}{n^{2}} = \frac{- 13.6 {(2)}^{2}}{n^{2}} = - 3.4$

n^{2} = 16

$n^{2} = 16$

∴ n = 4

$∴ n = 4$

n = 4, l = 2, m = 0

$n = 4, l = 2, m = 0$ belongs to

4 d_{y z}

$4 d_{y z}$ orbital

\to

$\to$ Radial nodes

\Rightarrow n - l - 1

$\Rightarrow n - l - 1$

\to

$\to$ Angular nodes

\Rightarrow 4 - 2 - 1 = 1

$\Rightarrow 4 - 2 - 1 = 1$

\to

$\to$ Angular nodes

\Rightarrow l \Rightarrow 2

$\Rightarrow l \Rightarrow 2$
Since

H e^{+}

$H e^{+}$ is a single

e^{-}

$e^{-}$ ion, its nuclear charge experienced.

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The ground state energy of hydrogen atom is -13.6 eV . Consider an electronic...

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