Question

An electron initially in the n = 5 orbital of a hydrogen atom emits a photon with a wavelength of 1284 nm. Calculate the

final orbital, nf.
O a. 6
Ob. 3
Ос. 2
O d. 1
e. 4

Answer 1

To determine the final orbital after an electron emits a photon, the Rydberg formula is applied. Given the initial orbital (n=5) and the emitted photon's wavelength (1284 nm), the calculations show that the final orbital is nf = 3.

Step-by-step explanation:

To determine the final orbital (nf) after an electron emits a photon from the n = 5 orbital, we need to use the Rydberg formula for hydrogen:

Rydberg equation: \( \frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \)

Here, \( \lambda \) is the wavelength of the emitted photon, R is the Rydberg constant (1.097 \times 10^7 m^-1), \( n_i \) is the initial energy level, and \( n_f \) is the final energy level.

Given data: \( n_i = 5 \) and \( \lambda = 1284 \mathrm{nm} \) or \( 1284 \times 10^{-9} m \).

Step 1: Convert the given wavelength to meters.

\( \lambda = 1284 \times 10^{-9} m \)

Step 2: Insert known values into the Rydberg formula and solve for \( n_f \).

\( \frac{1}{1284 \times 10^{-9} m} = 1.097 \times 10^7 m^-1 \left( \frac{1}{n_f^2} - \frac{1}{5^2} \right) \)

Step 3: Rearrange to solve for \( n_f^2 \) and then find \( n_f \).

After calculations, we find that the final orbital is n_f = 3.

This means the electron emitted a photon and transitioned from the \( n = 5 \) orbital to the \( n = 3 \) orbital.

Answer 2

Answer: 3

Step-by-step explanation:

You want to use the Rydberg equation which is given by

`E = -R_(H)((1)/(n_(final)^2) - (1)/(n_(initial)^2))`

where
`R_H = 2.18 \, \cdot 10^(-18) J`.

We are however given the wavelength instead of energy, and so we need to represent E in terms of its wavelength, and so we use
`E = (hc)/(\lambda)`. Remember that λ must be in meters in this formula. So we can divide 1284 nm by
`10^9` to get the wavelength in meters. This gives
`\lambda = 1.284 \, \cdot 10^(-6)` meters. Putting this all together we have

`(hc)/(\lambda) = -R_(H)((1)/(n_(final)^2) - (1)/(n_(initial)^2))`

where

`h = 6.626 \cdot 10^(-34) J\cdot s\\c = 2.998 \cdot 10^(8)\, m/s\\\lambda = 1.284 \cdot 10^(-6)\, m\\R_H = 2.18 \, \cdot 10^(-18) J\\n_(initial) = 5\\n_(final) = \, ?`

If we plug in everything we should get roughly
`n_f = 3.00194` but remember that orbitals must be a whole number and so you would round to the closest integer which gives
`n_f = 3.`