electrons emitting visible light

Question

What are the possible energy level changes for electrons emitting visible light in the hydrogen atom? Choose all that are correct.

1. n = 6 to n = 3

1. n = 5 to n = 3

1. n = 6 to n = 2

1. n = 3 to n = 1

1. n = 4 to n = 2

1. n = 4 to n = 1

1. n = 3 to n = 2

1. n = 4 to n = 1

1. n = 5 to n = 2

1. n = 2 to n = 1

Answer 1

We know that the energy difference between any two levels (first being higher than the second) follows the general rule (for hydrogen)...

E_np = -13.6 eV * (1/n² - 1/p²) = -2.178 x 10^-18 J * (1/n² - 1/p²)

where n is the upper shell and p is the lower shell.

We also know that E_np = hν with the knowledge that vλ = c ---> this gives us E_np = hc/λ

Rearranging for λ = hc/E_np = 1.986 x 10^-25 m³kg/s² / E_np

= 1.986 x 10^-25 m³kg/s² / [ -2.178 x 10^-18 J * (1/n² - 1/p²) ]

λ = -9.09 x 10^-8 / (1/n² - 1/p²) = -90.9 / (1/n² - 1/p²) in nanometers

Visible wavelengths of light account for ≈ 400 - 700 nm

Now we just need to compute the change in shell levels to determine the wavelengths generated

n p (1/n² - 1/p²) E_np (J) λ (nm) Visible

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6 3 -1/12 -1/12 * (-2.178 x 10^-18) = 1.815 x 10^-19 1098 no (IR)

5 3 -16/225 -16/225 * (-2.178 x 10^-18) = 1.549 x 10^-19 1278 no (IR)

6 2 -8/36 -8/36 * (-2.178 x 10^-18) = 4.84 x 10^-19 409 yes (indigo)

3 1 -8/9 -8/9 * (-2.178 x 10^-18) = 1.936 x 10^-18 102 no (UV)

4 2 -3/16 -3/16 * (-2.178 x 10^-18) = 4.08 x 10^-19 484 yes (greenish)

3 2 -5/36 -5/36 * (-2.178 x 10^-18) = 3.025 x 10^^-19 654 yes (red)

4 1 -15/16 -15/16 * (-2.178 x 10^-18) = 2.042 x 10^-18 97 no (UV)

5 2 -21/100 -21/100 * (-2.178 x 10^-18) = 4.57 x 10^^-19 433 yes (blue)

2 1 -3/4 -3/4 * (-2.178 x 10^-18) = 1.633 x 10^^-18 121 no (UV)

Answer 2

energy of an electron in hydrogen follows the formula:

-13.6 eV / n², where n is the energy level.

Therefore the energy from a transition is the difference between the two states:

1. ΔΕ = 13.6 eV (1/n_f² - 1/n_i²),

where n_f and n_i are the final and initial energy levels, respectively.

if you want to figure out what the wavelength of light when it's visible, you need to convert energy of a photo n into wavelength.

1. E = h*c / λ,

where h and c are plank's constant and the speed of light respectively. Then combine the two equations:

1. h*c / λ = 13.6 * (1/n_f² - 1/n_i²)

solve for lambda:

1. λ = h *c * / [13.6 * (1/n_f² - 1/n_i²)]

If you use h = 4.13 * 10^-15, and c = 3*10^8, you can plug in n_f and n_i for each equation to find the wavelength for all options.

If λ is between 400nm and 700nm (4*10^-7m and 7*10^-7m) the "choice is correct".