**– Select the correct answer from the following options:**

**– $(a) \ ^{14}C$**

**– $(b) \ ^{14}C^{2+}$**

**– $(c) \ ^{40}{\rm Ca}^{2+}$**

**– $(d) \ ^1H$**

**– $(e) \ ^{19}F^-$**

The aim of this article is to find the element in which the **number of electrons** is the **same** as the **number of neutrons**.

The basic concept behind this article is the **Subatomic Particles of an atom**.

An **atom** consists of **three Subatomic Particles**. The **Center of an Atom** or **Nuclei** has **Protons** and **Neutrons** whereas the **Electrons** are revolving around the **nuclei**. **Protons** are **positively charged particles** and **Electrons** are **negatively charged particles**. **Neutrons** are electrically **neutral**. For an electrically **neutral atom**, the number of **protons** is **equal** to the number of **electrons**.

\[Number\ of\ Protons=Number\ of\ Electrons\]

**Atomic Number** $Z$ of an element is the **number of protons** present in the **nuclei of its atom**.

**Mass Number** $A$ is the **sum** of the **number of protons** and **neutrons**.

\[Mass\ Number=Number\ of\ Neutrons+Number\ of\ Protons\]

Or:

\[Mass\ Number=Number\ of\ Neutrons+Atomic\ Number\ Z\]

\[A=N+Z\]

An **element** is scientifically represented by the following symbol:

\[{_Z^A}{\ X}^{\ q}\]

Where:

$A=$ **Mass Number**

$Z=$ **Atomic Number**

$q=$ **Electric Charge on an atom**

## Expert Answer

**Option (a) –** $\ ^{14}C$

**Atomic Number of Carbon** $C=6$

As per standard notation:

\[{_6^{14}}C\]

The **Number of Protons** and **Number of Electrons** is $6$ each.

**Mass Number** $A$ is $14$, so:

\[A=N+Z\]

\[14=N+6\]

\[N=8\]

Hence, the **Number of Electrons** is **not equal** to the **Number of Neutrons**.

**Option (b) –** $\ ^{14}C^{2+}$

**Atomic Number of Carbon** $C=6$

As per standard notation:

\[{_6^{14}}C^{2+}\]

The **Number of Protons** is $6$. Since the **atom** has a $2+$ **positive charge**, the **Number of Electrons** is:

\[E=6-2=4\]

**Mass Number** $A$ is $14$, so:

\[A=N+Z\]

\[14=N+6\]

\[N=8\]

Hence, the **Number of Electrons** is **not equal** to the **Number of Neutrons**.

**Option (c) –** $\ ^{40}{\rm Ca}^{2+}$

**Atomic Number of Calcium** $Ca=20$

As per standard notation:

\[{_{20}^{40}}{\rm Ca}^{2+}\]

The **Number of Protons** is $20$. Since the **atom** has a $2+$ **positive charge**, hence the **Number of Electrons** is:

\[E=20-2=18\]

**Mass Number** $A$ is $40$, so:

\[A=N+Z\]

\[40=N+20\]

\[N=20\]

Hence, the **Number of Electrons** is **not equal** to the **Number of Neutrons**.

**Option (d) –** $\ ^1H$

**Atomic Number of Hydrogen** $H=1$

As par standard notation:

\[{_1^1}H\]

The **Number of Protons** and **Number of Electrons** is $1$ each.

**Mass Number** $A$ is $1$, so:

\[A=N+Z\]

\[1=N+1\]

\[N=0\]

Hence, the **Number of Electrons** is **not equal** to the **Number of Neutrons**.

**Option (e)** – $\ ^{19}F^-$

**Atomic Number of Fluorine** $F=9$

As per standard notation:

\[{_9^{19}}F^-\]

The **Number of Protons** is $9$. Since the **atom** has a $-1$ **negative charge**, the **Number of Electrons** is:

\[E=9+1=10\]

**Mass Number** $A$ is $19$, so:

\[A=N+Z\]

\[19=N+9\]

\[N=10\]

Hence, the **Number of Electrons** is **equal** to the **Number of Neutrons**.

## Numerical Result:

**Option (e)** is correct. $\ ^{19}F^-$

## Example

Calculate the **Number of Electrons** and **Neutrons** in $\ ^{24}{\rm Mg}^{2+}$.

**Atomic Number of Magnesium** $Mg=12$

As par standard notation:

\[{_{12}^{24}}{\rm Mg}^{2+}\]

The **Number of Protons** is $12$. Since the **atom** has a $2+$ **positive charge**, the **Number of Electrons** is:

\[E=12-2=10\]

**Mass Number** $A$ is $24$, so:

\[A=N+Z\]

\[24=N+12\]

\[N=24\]