A single electron in an ion has ionization energy equal to `217.6eV`. What is the total number of neutrons present in one ion of it?

Options

A. (a) 2
B. (b) 4
C. (c) 5
D. (d) 9

Answer

To solve the problem, we need to determine the total number of neutrons present in an ion with a given ionization energy of 217.6 eV. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Ionization Energy
The ionization energy is the energy required to remove an electron from an atom or ion. In this case, we are given that the ionization energy is 217.6 eV.

### Step 2: Use Bohr's Theory
According to Bohr's theory, the energy of an electron in the nth orbit is given by the formula:
\[ E_n = -\frac{13.6 Z^2}{n^2} \]
where \(Z\) is the atomic number and \(n\) is the principal quantum number (the energy level of the electron).

### Step 3: Set Up the Equation
Since we are interested in the ionization energy, we can set the positive value of the ionization energy equal to the absolute value of the energy:
\[ 217.6 = \frac{13.6 Z^2}{1^2} \]
This simplifies to:
\[ 217.6 = 13.6 Z^2 \]

### Step 4: Solve for \(Z^2\)
To find \(Z^2\), we rearrange the equation:
\[ Z^2 = \frac{217.6}{13.6} \]
Calculating this gives:
\[ Z^2 = 16 \]

### Step 5: Find \(Z\)
Taking the square root of both sides, we find:
\[ Z = 4 \]
This indicates that the atomic number \(Z\) is 4.

### Step 6: Identify the Element
The atomic number \(Z = 4\) corresponds to the element beryllium (Be).

### Step 7: Determine the Mass Number
The most common isotope of beryllium has a mass number \(A = 9\).

### Step 8: Calculate the Number of Neutrons
The number of neutrons \(N\) can be calculated using the formula:
\[ N = A - Z \]
Substituting the values we have:
\[ N = 9 - 4 = 5 \]

### Final Answer
Thus, the total number of neutrons present in one ion of beryllium is **5**.

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