The decreasing order of wavelengths of the photons emitted or absorbed by the H atom is : b → c → a → d
Rydberg's formula used for calculating the wavelengths of the absorbed or emitted photons is given by,
,
where λ is the wavelength of the photon emitted or absorbed from an H atom electron transition from to and = 109677 is the Rydberg Constant. Here and represents the electron transitions of H atom.
(a) =2 to = infinity
= 109677/4 [since 1/infinity = 0] Therefore, = 4 / 109677 = 0.00003647 m
(b) =4 to = 20
= 6580.62
Therefore, = 1 / 6580.62 = 0.000152 m
(c) =3 to = 10
= 11089.56
Therefore, = 1 / 11089.56 = 0.00009 m
(d) =2 to = 1
= - 82257.75
Therefore, = 1 /82257.75 = - 0.0000121 m
[Even though there is a negative sign, the magnitude is only considered because the sign denotes that energy is emitted.]
So the decreasing order of wavelength of the photon absorbed or emitted is b → c → a → d.
The question is incomplete. The complete question is :
Arrange the following H atom electron transitions in order of decreasing wavelength of the photon absorbed or emitted:
(a) n = 2 to n = [infinity]
(b) n = 4 to n = 20
(c) n = 3 to n = 10
(d) n = 2 to n = 1
Learn more about the Rydberg's formula at brainly.com/question/14516049
#SPJ4