To determine the wavelength of light emitted by a hydrogen atom as its excited electron transitions from one energy level to another, we can use the Rydberg formula, which relates the wavelength of the emitted light (λ) to the initial and final energy levels of the electron. The formula is given as:
1/λ = R_H * (1/n_f^2 - 1/n_i^2),
where λ is the wavelength of the emitted light, R_H is the Rydberg constant for hydrogen (approximately 1.097 × 10^7 m⁻¹), n_f is the final energy level, and n_i is the initial energy level.
In this case, the initial energy level (n_i) is 5, and the final energy level (n_f) is 2. Plugging these values into the formula, we get:
1/λ = (1.097 × 10^7 m⁻¹) * (1/2^2 - 1/5^2).
Simplifying further:
1/λ = (1.097 × 10^7 m⁻¹) * (1/4 - 1/25), = (1.097 × 10^7 m⁻¹) * (21/100), = 2.3037 × 10^6 m⁻¹.
Taking the reciprocal of both sides:
λ = 1 / (2.3037 × 10^6 m⁻¹), = 4.345 × 10^-7 m.
Converting the wavelength to nanometers:
λ = 4.345 × 10^-7 m * 10^9 nm/m, = 434.5 nm.
Therefore, the wavelength of light emitted by a hydrogen atom as its excited electron falls from n=5 to n=2 is approximately 434.5 nm, not 23.037 × 10^9 nm as mentioned in the answer you provided.