**– Calculate the energy level of an electron in a hydrogen atom if it is considered to be in the ground state.**

The aim of this article is to find the **energy level of electrons** in a **hydrogen atom** when the hydrogen atom is in the **ground state** and **excited state**.

The basic concept behind this article is **Bohr’s theory of Energy Levels of Electrons**.

**Energy levels** **of electrons** are defined as the points where the electrons may exist having fixed distances from the nucleus of an atom. **Electrons** are **subatomic** particles that are **negatively** **charged**, and they **revolve** around the **nucleus** of an atom in a certain **orbit**.

For an atom having multiple **electrons**, these **electrons** are arranged around the **nucleus** in **orbits** in such a way that the **orbits** closest to the **nucleus** have **electrons** with **low energy** **levels**. These **Energy Level Orbits** are expressed as $n-level$, which are also called **Bohrâ€™s Orbits**.

As per **Bohrâ€™s Theory**, the equation for **energy level** is given by:

\[E=\frac{E_0}{n^2}\]

Where:

$E=$ **Energy Level of Electron in** $n^{th}$ **Bohrâ€™s Orbit**

$E_0=$ **Energy Level of Electron in the ground state**

$n=$ **Energy Level Orbits or Bohrâ€™s Orbit**

**Bohrâ€™s Theory** expressed the **energy levels** $n$ of a **hydrogen atom**, with the **first orbit** as **level-1** which is described as $n=1$ and defined as being the **ground state**. The **second orbit** called the **level-2** is expressed as $n=1$ and defined as the atom’s **first excited state**.

## Expert Answer

Given that we have a **hydrogen atom**, we need to find the **energy level** of the **electron** in a **hydrogen atom** when the **hydrogen atom** is in the **ground state** and **excited state** where:

\[n=4\]

As per **Bohrâ€™s Theory**, the **energy level** of the **electron** in $n^{th}$ **Bohrâ€™s Orbit** is expressed as follows:

\[E_n=\frac{E_0}{n^2}\]

We know that the **Energy Level of Electron** in the** ground state** $E_0$Â of the **hydrogen atom** is equal to:

\[E_0=-13.6eV\]

And for the **ground state**:

\[n=1\]

Substituting the values in the equation for **Bohrâ€™s Energy Level**:

\[E_1=\frac{-13.6eV}{{(1)}^2}\]

\[E_1=-13.6eV\]

As the units for **Energy** are usually **Joules** $J$, so **Electron Volt** $eV$ is converted to **Joules** as follows:

\[1eV=1.6\times{10}^{-19}J\]

So by converting the units:

\[E_1=-13.6\times(1.6\times{10}^{-19}J)\]

\[E_1=-21.76\times{10}^{-19}J\]

\[E_1=-2.176\times{10}^{-18}J\]

For the **excited** **state** of the **hydrogen** **atom**, we are given as:

\[n=4\]

Substituting the values in the above equation:

\[E_4=\frac{-13.6eV}{{(4)}^2}\]

\[E_4=-0.85eV\]

By converting the units from **Electron** **Volt** $eV$ to **Joules** $J$ as follows:

\[E_4=-0.85\times(1.6\times{10}^{-19}J)\]

\[E_4=-1.36\times{10}^{-19}J\]

## Numerical Result

The **energy level** of an **electron** in a **hydrogen** **atom** in the **ground state** is as follows:

\[E_1=-2.176\times{10}^{-18}J\]

The **energy level** of an **electron** in a **hydrogen** **atom** in an **excited state** at $n=4$ is as follows:

\[E_4=-1.36\times{10}^{-19}J\]

## Example

Calculate the **energy released** in a **hydrogen atom** when an **electron** **jumps** from $4^{th}$ to $2^{nd}$ **level**.

**Solution**

The **energy** that is **released** in a **hydrogen** **atom** when an **electron** **jumps** from $4^{th}$ to $2^{nd}$ **level** is calculated as follows:

\[E_{4\rightarrow2}=\frac{E_0}{{n_4}^2}-\frac{E_0}{{n_2}^2}\]

\[E_{4\rightarrow2}=\frac{(-13.6)}{{(4)}^2}-\frac{(-13.6)}{{(2)}^2}\]

\[E_{4\rightarrow2}=(-0.85eV)-(-3.4eV)\]

\[E_{4\rightarrow2}=2.55eV\]

By converting the units from **Electron** **Volt** $eV$ to **Joules** $J$ as follows:

\[E_{4\rightarrow2}=2.55\times(1.6\times{10}^{-19}J)\]

\[E_{4\rightarrow2}=4.08\times{10}^{-19}J\]