wavelength

Question

One of the emission lines for a one electron ion has a wavelength of 30.28nm for ni=4 and nf=2. The 253.4nm line has an initial state of 5. What is the quantum number of the final state for this line?

Answer 1

Use the relationship

 

1/λ = K(1/n_f² - 1/n_i²)

 

where K = constant, which we get from the given information.

 

1/30.28 = K(1/2² - 1/4²)

 

0.033025 = K(0.1875)

 

K = 0.17613

 

Then

 

1/253.4 = 0.17613(1/n_f² - 1/5²)

 

0.022406 = 1/n_f² - 1/25

 

Solve for n_f and get n_f = 4.

 

Answer 2

 First, you need to calculate the energy of the wave, using the formula E = hc/λ

 

where;

h is Planck's constant 6.626 x 10^-34 J*s

c is speed of light 3.0 x 10⁸ m/s.

λ  is wavelength in meters

 

Then you need to plug the energy into the Rydberg formula:

 

E =RZ²(1/(n_final)² - 1/(n_initial)²)

 

where n is the energy level of the electron, Z is the atomic number and R is the Rydberg constant. 

 Since we don't know what R or Z is in this problem. Let X = RZ² and solve for it.

 

E = X(1/nf² - 1/ni²)

 

Then you can use your newly acquired X value to solve for nf in the second part of the question.