Question

the energy of a singly ionized LI atom in the he first excited state apporximately
​

Answer 1

Explanation:

From the Bohr's theory of single electron species, we have the Energy of the state (quantum number = n) and atomic number = Z as

E=−13.6

n

2

Z

2

Thus, Energy of 4

th

state in Helium ion is E

4

He

+

=−13.6×

4

2

2

2

=−3.4eV

Energy of the emitted photon is E

p

=2.6eV

Energy after emitting the photon is E

f

=E

4

He

+

−E

p

=−3.4−2.6=−6eV

The quantum number is given by n=(

E

n

−13.6×Z

2

)

2

1

=

−6

−13.6×4

=

9

=3

Answer 2

Answer:

The energy of a singly ionized LI atom in the first excited state is approximately Equal to E =−30.6eV/atom

​Explanation:

From the Bohr's theory of single-electron species, we have the Energy of the state (quantum number = n) and atomic number = Z as

E=−13.6 $\frac{Z^{2} }{n^{2} }$ eV/atom.

Li has 3 electrons.  Li ^2+ has one electron. It is a hydrogen-like atom. In the first excited state, the electron is in the 2s orbital.

Its energy is  E=−13.6 $\frac{Z^{2} }{n^{2} }$ eV/atom.

​here Z=3,n=2

​ E=−13.6 $\frac{3^{2} }{2^{2} }$

E =−30.6eV/atom