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One of the emission lines of the hydrogen atom has a wavelength of 93.07 nm. Determine the initial and final values of n associated with this emission.

The question aims to find the initial and final values of n when the wavelength of emission of the hydrogen line is 93.07. 

Distance between the same points (adjacent peaks) in an adjacent cycle of a waveform signal propagating in space or along a line is called wavelength. For wireless systems, this length is usually expressed in meters (m), centimeters (cm), or millimeters (mm). For infrared (IR), visible light, ultraviolet (UV), and gamma rays (γ), the wavelength is more commonly specified in nanometers (nm).

Mathematical Representation:

Traveling sinusoidal wave represents as follows:

\[y(t)=Acos(\dfrac{2\pi}{\lambda}(x-vt)\]

$y$ in the equation is the value of the wave at each position $x$ and time $t$, and $A$ is the amplitude of the wave. They are also generally expressed in wavenumber $k$ ($2\pi$ times the reciprocal of the wavelength) and angular frequency $ω$ ($2\pi$ times the frequency) as follows:

\[y(x,t)=A\cos(kx-\omega t)=A\cos(k(x-vt))\]

\[k=\dfrac{2\pi}{\lambda}=\frac{\omega}{v}\]

or

The formula for wavelength is given by:

\[\lambda=\dfrac{v}{f}\]

where $v$ is the velocity and $f$ is the frequency.

Expert Answer

Wavelength $\lambda$ for emission line is given by:

\[\lambda=93.07nm=93.07\times 10^-9 m\]

This wavelength is associated with the ultraviolet region of the electromagnetic spectrum.

Rydberg constant is given by:

\[R_{H}=1.0967\times 10^7 m^{-1}\]

The final value of the $n$ associated with the emission is calculated as

Lyman series is used to calculate the final value for the wavelength of the emission spectrum of the hydrogen line.

For the Layman series of an emission line of   $n_{f}=1$.

The initial value of the $n$ associated with the emission is calculated as:

\[\dfrac{1}{\lambda}=(R_{H})(\dfrac{1}{n_{f}^2}-\dfrac{1}{n_{i}^2})\]

\[-\dfrac{1}{n_{i}^2}=\dfrac{1}{\lambda\times R_{H}}-\dfrac{1}{n_{f}^2}\]

Plug the values of $R_{H}$ and $n_{f}$ into the above equation:

\[-\dfrac{1}{n_{i}^2}=\dfrac{1}{93.07\times 10^{-9}m\times1.09677\times 10^7 m^{-1}}\]

\[-\dfrac{1}{n_{i}^2}=0.02\]

\[n_{i}=7\]

Numerical Results

The initial value of the $n$ associated with the emission is :

\[n_{i}=7\]

The final value of the $n$ associated with the emission is:

\[n_{f}=1\]

Example

One of the emission lines from a hydrogen atom has a wavelength of $120 nm$. Calculate the final and initial values of $n$ associated with this release.

Solution:

Wavelength $\lambda$ for emission line is given by:

\[\lambda=120nm=120\times 10^-9 m\]

This wavelength is associated with the ultraviolet region of the electromagnetic spectrum.

Rydberg constant is given by:

\[R_{H}=1.0967\times 10^7 m^{-1}\]

The final value of the $n$ associated with the emission is calculated as

Lyman series is used to calculate the final value for the wavelength of the emission spectrum of the hydrogen line.

For the Layman series of an emission line of   $n_{f}=1$.

The initial value of the $n$ associated with the emission is calculated as:

\[\dfrac{1}{\lambda}=(R_{H})(\dfrac{1}{n_{f}^2}-\dfrac{1}{n_{i}^2})\]

\[-\dfrac{1}{n_{i}^2}=\dfrac{1}{\lambda\times R_{H}}-\dfrac{1}{n_{f}^2}\]

Plug the values of $R_{H}$ and $n_{f}$ into the above equation:

\[-\dfrac{1}{n_{i}^2}=\dfrac{1}{120\times 10^{-9}m\times1.09677\times 10^7 m^{-1}}\]

\[-\dfrac{1}{n_{i}^2}=0.759\]

\[n_{i}=1.147\]

The initial value of the $n$ associated with the emission is calculated as:

\[n_{i}=1.147\]

The final value of the $n$ associated with the emission is calculated as:

\[n_{f}=1\]

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