Question

An electron in the n=6 level of the hydrogen atom relaxes to a lower energy level, emitting light of ?=93.8nm. find the principal level to which the electron relaxed.

Answer 1

We can find the principal level or lower level using Rydberg's formula:

1/w = R(1/L² - 1/U²) 

where:

w is the wavelength (93.8 nm),

L is the lower energy level (unknown)

U the upper energy level (n= 6)

R is Rydberg's constant (10,967,758 waves per meter)

Substituing known values into the equation:
1/(9.38 * 10^-8 m.) = 10,967,758(1/L² - 1-36)

Using the solver function of the calculator to get for L:

L = 0.999 
so L = 1. 
The lower level is 1 (the ground state).

Answer 2

The electron in the hydrogen atom relaxes to $\boxed{\text{first}}$ principal level.

Further Explanation:

Electronic transition is the process of transference of electrons from one energy level to another energy level. If electron jumps from higher to lower level, it is accomplished by release of energy. This process is called emission process. Similarly, electron gains energy while going from lower to higher energy levels. This process is known as absorption process.

The expression to calculate wavelength of transition in hydrogen atom is as follows:

$\dfrac{1}{\lambda}=R_\text{H}\left(\dfrac{1}{\text{n}_1^2}-\dfrac{1}{{\text{n}_2^2}}\right)$                                                            ...... (1)                                        

Where,

$\lambda$  is the wavelength of transition.

$R_\text{H}$ is Rydberg constant.

$\text{n}_2$ is the initial energy level of transition.

$\text{n}_1$ is the final energy level of transition.

Rearrange equation (1) to calculate $\text{n}_1$.

$\dfrac{1}{\text{n}_2}=\sqrt{\dfrac{1}{\text{n}_1^2}-\dfrac{1}{\lambda R}_\text{H} }$                                                  ...... (2)

Substitute 6 for $\text{n}_1$, 93.8 nm for $\lambda$ and $1.0974\times10^7\text{ m}^{-1}$ for $R_\text{H}$ in equation (2).

$\begin{aligned}\dfrac{1}{\text{n}_2}&=\sqrt{\dfrac{1}{\text{6}^2}-\dfrac{1}{(93.8\text{ nm})\left(\dfrac{10^{-9}\text{ m}}{1\text{ nm}}\right)(1.0974\times10^7\text{ m}^{-1})}\\&=0.999\\&\approx{1}\end{aligned}$                                              

Solving for $\text{n}_2$,

$\text{n}_2=1$

Therefore electron present in sixth energy level relaxes to first energy level.

Learn more:

1. Which transition is associated with the greatest energy change? https://brainly.com/question/1594022

2. Describe the spectrum of elemental hydrogen gas: https://brainly.com/question/6255073

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Atomic structure

Keywords: electronic transition, emission, absorption, lower, higher, energy level, electrons, wavelength, Rydberg constant, initial energy level, final energy level, sixth, first, n1, n2, 0.999, 1, wavelength of transition, hydrogen atom, higher, lower.