Show that the various lines in the hydrogen spectrum can be expressed in angstroms as λ(A) = 911n where ni 1 for the Lyman series, 2 for the Balmer series, and 3 for the Paschen series The integer n is larger than nı. (20 points) (b) Calculate for the Lyman series to n-5, the Balmer series to n 7, and the Paschen series to n 10. Plot the results as in Fig. 2-2 of the textbook. What are the wavelength limits for each of the three series?

Respuesta :

Answer:

Explanation:

It is spectrum of the hydrogen atom is well known the first lines of the Lyman series are: 1216, 1025, 972, 949, 937 A

These series of increasingly close lines were studied and found from the equation c = L F the frequency varies with the inverse of the wavelength

The series has the general form of convergence towards a puno in the infinite in the Lyman series this point is about 919 A, so this convergence is explained by a mathematical series of that of some number that is the end of the series power of the 1/xⁿ form;

Let's try to find the explicit series, let's find the inverse of each wavelength

λ           1/λ              1/λ (x1000)      n      (1- 1/n)    (1- 1/n²)

1216    0.000822    0.822            2        0.5        0.75

1025   0.0009756  0.976            3        0.66      0.889

972     0.00103       1.030             4        0.75      0.9375

949     0.00105      1.050              5        0.8        0.96

937     0.001067    1.067              6        0.833     0.972

Let us graph each case where the axis and place the Lym series values, multiplied by 1000 for clarity and each series, from the graph examination it can be seen that the series with 1 / n2 is better suited to the experimental values

Therefore the functional form of the series is

        1 /λ = C (1 -1 / n²)

Let's perform the same procedure for the other series, finding the correct functional form is

Balmer     1 /λ = C2 (1/2² -1/n²)

Paschem 1 /λ = C3 (1/3² - 1/n²)

After seeing these three series we can propose an eral form for all the series of the different lines

 1 / λ = R (1 /n₀² - 1 / Nₓ²)

 R = 1,097 107 m-1

With nₓ > n₀

b) let's calculate the wavelength for each point

Lymas n = 5

                1 / L = 1.097 107 (1 - 1/52) = 1.097 107 0.96

                1 / L = 1.05 107

                 L = 9495.5 A

Balmer n = 7

                1 / L = 1.097 107 (1/22 - 1/72) = 1.097 107 0.229

                 1 / L = 0.2518 107

                   L = 3970.4 A

Paschem n = 10

        1 / L = 1,097 107 (1/32 - 1/102) = 1,097 107 0.1011

  1 / L = 11091.9 A