Question

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

Answer 1

We can find the lower level by using Rydberg's formula: 1/w = R(1/L² - 1/U²) where:

w - wavelength (9.38nm),

L - lower energy level

U - upper energy level (6); and

R - Rydberg's constant (10,967,758 waves per meter for hydrogen).

So plugging in the information above: 

1/ (9.38 * 10**-8 m.) = 10,967,758(1/L² - 1-36)

But we need to solve for L. 

The left side (1/9.38 * 10**-8) becomes 10,660,980

So translating:

10660980 = 10967758(1/L² - 0.02777777)

Then divide both sides by 10967758
we get: 0.9720291 = 1/L² - 0.02777777.

then add 0.02777777 to both sides 

we get: 0.9998 = 1/L² 

0.9998 is very close to 1, so we can approximate 1/L² = 1 so L = 1. 


Therefore the lower level is 1 which is the ground state.

Answer 2

The principal energy level to which the electron relaxed is $\boxed{n=1}$ .

Further Explanation:

When the transition of the electron from one energy level to another energy level takes place, there is an emission of the particular wavelength according to the energy of the level. The relation between the energy of emission and the wavelength is given as:

$E=\frac{{hc}}{\lambda }$

Substitute the value of the Planck’s constant, speed of light and the wavelength of the emitted light in the above expression.

$\begin{aligned}E&=\frac{{\left({6.63\times{{10}^{-34}}}\right)\times\left({3\times{{10}^8}}\right)}}{{93.8\times{{10}^{-9}}}}\\&=2.12\times{10^{-18}}\,{\text{J}}\\\end{aligned}$

This amount of energy is the change in the energy level as the electronic transition takes place.

The energy of the different energy levels in a hydrogen atom is expressed as:

$\Delta E=13.6\,{\text{eV}}\left({\frac{1}{{n_f^2}}-\frac{1}{{n_i^2}}}\right)$

Here, $\Delta E$  is the change in the energy, ${n_f}$  is the final energy level and ${n_i}$  is the initial energy level.

Substitute the values in the above expression.

$\begin{aligned}2.12\times{10^{-18}}&=13.6\times\left({1.6\times{{10}^{-19}}}\right)\left({\frac{1}{{n_f^2}}-\frac{1}{{{6^2}}}}\right)\\\frac{{2.12\times{{10}^{-18}}}}{{2.176\times{{10}^{-18}}}}&=\frac{1}{{n_f^2}}-\frac{1}{{36}}\\\frac{1}{{n_f^2}}&=0.97+\frac{1}{{36}}\\{n_f}&\approx 1\\\end{aligned}$

Thus, the principal energy level to which the electron relaxed is $\boxed{n=1}$ .

Learn More:

1. Microwave ovens emit microwave energy with a wavelength of 12.5 cm. What is the energy of exactly one photon of this microwave radiation brainly.com/question/2385939

2. What is the kinetic energy of the emitted electrons when cesium is exposed to uv rays of frequency 1.9×1015hz brainly.com/question/9059731

3. When the atom's electrons step down to lower energy levels in a thin cloud of hot gas, what is produced brainly.com/question/6054540

Answer Details:

Grade: High School

Subject: Physics

Chapter: Hydrogen spectrum

Keywords:

Electron, level n=6, hydrogen atom, transition of electron, lower energy level, emitting light, wavelength, 93.8 nm, principal level, electron relaxed.