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Calculate the wavelength of the photon emitted when an electron makes a transition from n=4 to n=3. You can make use of the following constants:

h=6.626×10^−34 J⋅s

c=2.998×10^8 m/s

1 m=10^9 nm


Express your answer to four significant figures and include the appropriate units.

Respuesta :

Iqta

Answer:  

1875 nm

Explanation:

As per the data given:

                                  Initial orbit of electron = ni = 4

                                  Final orbit of electron = nf = 3.

As we know that,

                           Speed of electromagnetic waves = c = 3 × 10^8 m/s

                           Planck’s constant = h = 6.63 × 10^−34 Js.  

We will use Rydberg’s equation to get the wavelength or λ.

                                          1/λ=R (1/nf^2 – 1/ni^2)

Here, R= Rydberg Constant 1.0974x10^7 m-1;  λ is the wavelength;  n is equal to the energy level (initial and final)

                                           1/λ=R (1/nf^2 – 1/ni^2)

                                           1/λ =1.0974x10^7 (1/3^2 – 1/4^2)

                                           1/λ =1.0974x10^7 (1/9 – 1/16)

                                           1/λ =1.0974x10^7 (0.0486)

                                            1/λ=0.05333x 10^7

                                             λ= 1/ 0.05333x 10^7

                                             λ =1.875 x10^ -6m

Converting λ =1.875 x10^-6m to nanometers (nm )

                                               As, 1 m=10^9 nm                                  

                                               λ =1.875 x10^-6m x 10^9

                                                λ =1875 nm  

Note:

The value 1.0974x10^7 in the Rydberg equation comes from the Bohr equation and is equal to delta E / hc:  

delta E / hc = (2.18 x 10^-18 J) / (6.626 x 10^-34 J s)(3.00 x 10^8 m/s) = 1.0967 x 10^7

Hope it helps! :)


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