Question

The energy of the first excited state of Li2+ . Pls help and solve the answer

Answer 1

Answer:

The energy of the first excited state of $Li^{2+}$ is -30.6 eV.

Explanation:

The energy of any energy level or state of hydrogen element is given by,

$E_n = -(\frac{me^4}{8\epsilon_0^2 h^2} )(\frac{Z^2}{n^2} )$                    --------------------------  (i)

Where $E_n$ = energy

Z = Atomic number

n = Principal quantum number

Solving $(\frac{me^4}{8\epsilon_0^2 h^2} )$, we get

$(\frac{me^4}{8\epsilon_0^2 h^2} ) = 13.6$

Now, (i) becomes

$E_n = -13.6 (\frac{Z^2}{n^2} )$                    --------------------- (ii)

Given element is lithium (Li)

Atomic number of lithium, Z = 3

It means that it has 3 electrons. So  $Li^{2+}$ has only 1 electron and it will behave like hydrogen.

In the first excited state of  $Li^{2+}$, electron will jump from 1s to 2s orbital.

For 2s, principal quantum number, n = 2

Putting the values in equation (ii), we get

$E_2 = -13.6(\frac{3^2}{2^2} )\\\\E_2 = -13.6(\frac{9}{4} )\\\\E_2 = -13.6 \times 2.25 \ eV \\\\E_2 = -30.6 eV$

Hence, the energy of the first excited state of $Li^{2+}$ is $E_2$ = -30.6 eV.

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Answer 2

Answer: The initial excited state of has an energy of -30.6 eV.

Explanation:

The energy of every hydrogen element's energy level or state is given by, (i)Energy = where

Atomic number Z

n = Primary quantum number

We obtain by solving

I has now be The element in question is lithium (Li)

Lithium's atomic number is Z = 3.

It signifies it contains three electrons. So it just has one electron and behaves like hydrogen.

The electron will hop from the 1s to the 2s orbital in the first excited state of.n = 2 for the primary quantum number 2s

When we plug the data into equation (ii), we obtain

As a result, the energy of the initial excited state is = -30.6 eV.

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