Determine the electronic transition in
Li+2
ion
emitting lights of wavelength equal to the
wavelength of 2nd Balmer line (n2=4 to n1=1)
of
He
ion.
Answers
Explanation:
According to Bohr's Theory, the energy of an electron in n
th
orbit is:
E
n
=−R
H
(
n
2
Z
2
)J
where R
H
=2.18×10
−18
J and Z is the atomic number of an atom or ion having one electron.
In the atomic emission spectra of Hydrogen, the electron jumps from higher orbit/ energy state n
2
to lower energy state n
1
with the emission of radiation. The energy of the emitted radiation is equal to the difference between the energy of the two states involved:
ΔE=E
n
1
−E
n
2
Therefore, solving for first line of Balmer series, n
1
=2andn
2
−n
1
=1 for Hydrogen, Z=1 energy of emitted radiation is:
ΔE=−R
H
(
2
2
1
−
3
2
1
)
For Li
2+
(Z=3), energy of radiation is:
ΔE=−R
H
.3
2
(
n
1
2
1
−
n
2
2
1
)
For the two radiations to be of same wavelength, energy difference between transition states should be same. Hence,
1
2
(
2
2
1
−
3
2
1
)=3
2
(
n
1
2
1
−
n
2
2
1
)
Or,
2
2
1
=
n
1
2
3
2
and
3
2
1
=
n
2
2
3
2
Hence,
n
1
=6andn
2
=9 transition is the answer.