Respuesta :
The decreasing order of wavelengths of the photons emitted or absorbed by the H atom is : b → c → a → d
Rydberg's formula :
[tex]\frac{1}{\lambda} = R_h (\frac{1}{n_1^2}-\frac{1}{n_2^2} )[/tex],
where λ is the wavelength of the photon emitted or absorbed from an H atom electron transition from [tex]n_1[/tex] to [tex]n_2[/tex] and [tex]R_h[/tex] = 109677 is the Rydberg Constant. Here [tex]n_1[/tex] and [tex]n_2[/tex] represents the transitions.
(a) [tex]n_1[/tex] =2 to [tex]n_2[/tex] = infinity
[tex]\frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{\infty^2} )[/tex] = 109677/4 [since 1/infinity = 0] Therefore, [tex]\lambda[/tex] = 4 / 109677 = 0.00003647 m
(b) [tex]n_1[/tex]=4 to [tex]n_2[/tex] = 20
[tex]\frac{1}{\lambda} = 109677\times (\frac{1}{4^2}-\frac{1}{20^2} )[/tex] = 6580.62
Therefore, [tex]\lambda[/tex] = 1 / 6580.62 = 0.000152 m
(c) [tex]n_1[/tex]=3 to [tex]n_2[/tex] = 10
[tex]\frac{1}{\lambda} = 109677\times (\frac{1}{3^2}-\frac{1}{10^2} )[/tex] = 11089.56
Therefore, [tex]\lambda[/tex] = 1 / 11089.56 = 0.00009 m
(d) [tex]n_1[/tex]=2 to [tex]n_2[/tex] = 1
[tex]\frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{1^2} )[/tex] = - 82257.75
Therefore, [tex]\lambda[/tex] = 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.
Learn more about the Rydberg's formula athttps://brainly.com/question/14649374
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