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:

\[n_{f}=1\]