Question

Arrange the following H atom electron transitions in order of decreasing wavelength of the photon absorbed or emitted:

Answer 1

The decreasing order of wavelengths of the photons emitted or absorbed by the H atom is : b → c → a → d

Rydberg's formula :

                                   $\frac{1}{\lambda} = R_h (\frac{1}{n_1^2}-\frac{1}{n_2^2} )$,

where  λ is the wavelength of the photon emitted or absorbed from an H atom electron transition from $n_1$ to $n_2$ and $R_h$ = 109677 is the Rydberg Constant. Here $n_1$ and  $n_2$ represents the transitions.

(a) $n_1$ =2 to $n_2$ = infinity

            $\frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{\infty^2} )$ = 109677/4     [since 1/infinity = 0] Therefore, $\lambda$ = 4 / 109677 = 0.00003647 m

(b) $n_1$=4 to  $n_2$ = 20

           $\frac{1}{\lambda} = 109677\times (\frac{1}{4^2}-\frac{1}{20^2} )$ = 6580.62

Therefore,  $\lambda$ = 1 / 6580.62 = 0.000152 m

(c) $n_1$=3 to  $n_2$ = 10

          $\frac{1}{\lambda} = 109677\times (\frac{1}{3^2}-\frac{1}{10^2} )$ = 11089.56

Therefore,  $\lambda$ = 1 / 11089.56 = 0.00009 m

(d)  $n_1$=2 to  $n_2$ = 1

          $\frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{1^2} )$ = - 82257.75

Therefore,  $\lambda$ = 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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