Question

Calculate the frequency of the n = 6 line in the Lyman series of hydrogen.

Answer 1

The frequency of the n = 6 line in the Lyman series of hydrogen is $\boxed{{\mathbf{3}}{\mathbf{.196 \times 10 }}{{\mathbf{s}}^{{\mathbf{ - 1}}}}}$.

Further Explanation:

The given problem is based on the concept of the emission spectrum of a hydrogen atom.

Lyman series:

When an electron undergoes transition from any higher energy orbit $\left({{{\text{n}}_{\text{f}}}=2,{\text{ }}3,{\text{ }}4,...}\right)$ to first energy orbit $\left({{{\text{n}}_{\text{i}}} = 1}\right)$ then it emits the energy to complete the process. This spectral lines formed due to this emission is known as the Lyman series of hydrogen atom.

The formula to calculate the energy of transition in the hydrogen atom is,

$\Delta E={R_{\text{H}}}\left({\frac{1}{{{{\left({{{\text{n}}_{\text{i}}}}\right)}^2}}}-\frac{1}{{{{\left({{{\text{n}}_{\text{f}}}}\right)}^2}}}}\right)$

Where,

$\Delta E$ is the energy difference between two energy levels.

${R_{\text{H}}}$ is a Rydberg constant and its value is $2.179 \times {10^{ - 18}}{\text{ J}}$.

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

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

Substitute the 1 for ${{\text{n}}_{\text{i}}}$ , 6 for ${{\text{n}}_{\text{f}}}$ and $2.179\times {10^{ - 18}}{\text{ J}}$ for ${{\text{R}}_{\text{H}}}$ in the above formula to calculate the value of energy of the given transition.

$\begin{aligned}\Delta E &={R_{\text{H}}}\left({\frac{1}{{{{\left( {{{\text{n}}_{\text{i}}}}\right)}^2}}} \frac{1}{{{{\left({{{\text{n}}_{\text{f}}}}\right)}^2}}}}\right)\\&=\left({2.179 \times {{10}^{ - 18}}{\text{ J}}} \right)\left({\frac{1}{{{{\left( {\text{1}}\right)}^2}}}-\frac{1}{{{{\left({\text{6}} \right)}^2}}}}\right)\\&={\mathbf{2}}{\mathbf{.118 \times 1}}{{\mathbf{0}}^{{\mathbf{ - 18}}}}{\mathbf{ J}}\\\end{aligned}$

Now we can calculate the frequency of the given transition by using the following formula.

$\Delta E = hv$

Where,

$\Delta E$ is the energy difference between two energy levels.

$h$ is a Plank’s constant and its value is $6.626\times {10^{ - 34}}{\text{ Js}}$.

v is a frequency of the transition.

Rearrange the above formula to calculate the frequency of the given transition and substitute the value of $\Delta E$.

$\begin{aligned}\Delta E&=hv\hfill\\v&=\frac{{\Delta E}}{h}\hfill\\&=\frac{{{\text{2}}{\text{.118}} \times {\text{1}}{{\text{0}}^{ - {\text{18}}}}{\text{ J}}}}{{6.626 \times {{10}^{ - 34}}{\text{ Js}}}} \hfill\\&={\mathbf{3}}{\mathbf{.196 \times 10 }}{{\mathbf{s}}^{{\mathbf{ - 1}}}} \hfill \\\end{aligned}$

Learn more:

1. Calculation of oxidation state: brainly.com/question/2086855

2. Determine how many moles of water produce: brainly.com/question/1405182

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Structure of atom

Keywords: Lyman series of hydrogen, hydrogen, n=6 transition, emission spectra, Plank’s constant, transition, energy orbits.

Answer 2

Lyman Series Working Formula:

1/λ = RH (1-(1/n^2))

Given:

n = 6
RH = Rydberg's constant = 1.0968x10^7 m^-1
c = speed of light = 3x10^8 m/s

Required: 

Frequency (Hertz or cycles per second)

Solution:

To solve for the wavelength λ, we substitute the given in the working formula
1/λ = RH (1-(1/n^2))
1/λ = 1.0968x10^7 m^-1 (1-(1/6^2))
λ = 0.0000000938 m or 93.8 nm 

To get the frequency, we will use the formula below. 
f = c/λ

We then substitute c or the speed of light,
f = (3x10^8 m/s) / 0.0000000938 m 

Therefore,
f = 3.2x10^15 s^-1

ANSWER: Frequency = 3.2x10^15 s^-1