Question

For hydrogen, what is the wavelength of the photon emitted when an electron drops from a 4d orbital to a 2p orbital in a hydrogen atom? The Rydberg constant is 1.097 × 10-2 nm-1.

Answer 1

Answer:

486.17 nm

Explanation:

Using the formula  $\frac{1}{\beta } = R__H (\frac{1}{n_1^2} - \frac{1}{n_2^2})$ to calculate for the wavelength of the photon emitted. where beta means a substitute for  wavelength (λ) in the above equation.

Our given parameters include ;

The Rydberg constant is 1.097 × 10-2 nm-1.

an electron drops from a 4d orbital to a 2p orbital; which implies the n₂ & n₁ respectively.

Substituting them to the above equation; we can calculate for our wavelength easily;

∴

$\frac{1}{\beta}=1.097*10^{-2}(\frac{1}{2^2} - \frac{1}{4^2})$

$\frac{1}{\beta}=1.097*10^{-2}*0.1875$

$\frac{1}{\beta}=0.00205875$

Since we've earlier said our (β) serves as a substitute for wavelength (λ)

Thus; λ = 0.002056875⁻¹

λ = 486.174 nm

λ ≅ 486.17 nm