Question

Calculate the wavelength of the light emitted when an electron in a hydrogen atom makes each transition and indicate the region of the electromagnetic spectrum (infrared, visible, ultraviolet, etc.) where the light is found. n = 4 → n = 2

Answer 1

The wavelength of the light emitted when an electron transitions from n = 4 to n = 2 in a hydrogen atom is approximately 2.84 x 10^-7 meters. This light falls in the visible region of the electromagnetic spectrum.

Step-by-step explanation:

To calculate the wavelength of the light emitted when an electron in a hydrogen atom makes a transition from n = 4 to n = 2, we can use the Rydberg formula:

1/λ = R_H (1/n_f^2 - 1/n_i^2)

Here, R_H is the Rydberg constant for hydrogen (1.097 x 10^7 m^-1), n_f is the final energy level (2), and n_i is the initial energy level (4).

Plugging in the values and solving for λ, we get:

1/λ = (1.097 x 10^7 m^-1) (1/4^2 - 1/2^2)

1/λ = (1.097 x 10^7 m^-1) (1/16 - 1/4)

1/λ = (1.097 x 10^7 m^-1) (3/16)

1/λ = (3.516 x 10^6 m^-1)

Therefore, the wavelength (λ) of the light emitted is approximately 2.84 x 10^-7 meters. This light falls in the visible region of the electromagnetic spectrum.