Calculate the shortest and longest wavelength in case of hydrogen atom in lymann series
Answers
Answer:
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Explanation:
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ANSWER
For Lyman series, n
1
=1.
For shortest wavelength in Lyman series (i.e., series limit), the energy difference in two states showing transition should be maximum, i.e., n
2
=∞.
So,
λ
1
=R
H
[
1
2
1
−
∞
2
1
]=R
H
λ=
109678
1
=9.117×10
−6
cm
=911.7
A
˚
For longest wavelength in Lyman series (i.e., first line), the energy difference in two states showing transition should be minimum, i.e., n
2
=2.
So,
λ
1
=R
H
[
1
2
1
−
2
2
1
]=
4
3
R
H
or λ=
3
4
×
R
H
1
=
3×109678
4
=1215.7×10
−8
cm
=1215.7
A
˚
.
Answer:
According to Balmer formula
ṽ=1/λ = RH[1/n12-1/n22]
For the Balmer series, ni = 2.
Thus, the expression of wavenumber(ṽ) is given by,
Wave number (ṽ) is inversely proportional to wavelength of transition.
Hence, for the longest wavelength transition, ṽ has to be the smallest.
For ṽ to be minimum, nf should be minimum.
For the Balmer series, a transition from ni = 2 to nf = 3 is allowed.
Hence, taking nf = 3,we get:
ṽ= 1.5236 × 106 m–1
Explanation: