Question

The wavelength of the radiation emitted, when a Hydrogen atom electron falls from infinity to the stationary state, would be:

Rydberg constant = 1.097 x 10⁷ m⁻¹

Answer 1

The wavelength of the radiation emitted when a Hydrogen atom electron transitions from infinity to the ground state is found by using the Rydberg formula, yielding a wavelength in the ultraviolet spectrum approximately equal to 9.11 x 10⁻¸ meters.

Step-by-step explanation:

The wavelength of the radiation emitted when a Hydrogen atom electron falls from infinity to the stationary state can be calculated using the Rydberg formula. Given the value of the Rydberg constant (R) as 1.097 x 10⁷ m⁻¹, and knowing that an electron falling from infinity falls to the first stationary state (n = 1) since it's the ground state of the hydrogen atom, we can apply this concept to find the wavelength of the emitted electromagnetic (EM) radiation.

According to the Rydberg formula for hydrogen:

1/λ = R (1/n²1 - 1/n²2)

For an electron transitioning from infinity (where the energy level is considered to be 0), let's take n2 as ∞, which makes the second term 0. So, for n1 = 1 (ground state), the equation simplifies to:

1/λ = R (1/1²)

λ = 1/R

λ = 1/(1.097 x 10⁷ m⁻¹)

λ = 9.11 x 10⁻¸ m

This calculation shows that the wavelength corresponds to light in the ultraviolet spectrum. The exact value may be slightly different due to the discrete quantum nature of energy transitions in an atom, but this provides a close approximation based on the provided Rydberg constant.