Respuesta :
The order of increasing frequency of the photon absorbed or emitted by the H atom is :
d → a → c → b.
The wavelength is calculated using the Rydberg's formula given by,
[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.
Now the frequency, [tex]f=\frac{c}{\lambda}[/tex] , where c = [tex]3\times 10^8 ms^{-1}[/tex] is the speed of light.
(a) [tex]n_1[/tex] =2 to [tex]n_2[/tex] = 4
[tex]\frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{4^2} )[/tex] = 20564.4375 [since 1/infinity = 0] Therefore, f =c/ [tex]\lambda[/tex] = 61693.3125 x[tex]10^8[/tex] Hz
(b) [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, f=c/ [tex]\lambda[/tex] = 246773.25 x [tex]10^8[/tex] Hz
(c) [tex]n_1[/tex]=2 to [tex]n_2[/tex] = 5
[tex]\frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{5^2} )[/tex] = 23032.17
Therefore, f=c/[tex]\lambda[/tex] = 69096.51 x [tex]10^8[/tex] Hz
(d) [tex]n_1[/tex]=4 to [tex]n_2[/tex] = 3
[tex]\frac{1}{\lambda} = 109677\times (\frac{1}{4^2}-\frac{1}{3^2} )[/tex] = -5331.52
Therefore, f =c/[tex]\lambda[/tex] = - 15994.56x [tex]10^8[/tex] Hz
We compare the frequencies considering only the magnitudes of the frequency. Because the sign of the frequency implies whether a photon is emitted or absorbed.
Thus the increasing order of frequencies of the photon absorbed or emitted in H atom electron transitions is d → a → c → b.
Learn more about the Rydberg's formula at https://brainly.com/question/14649374
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