According to Bohr's model, if the kinetic energy of an electron in `2^(nd)` orbit of `He^(+)` is `x`, then what should be the ionisation energy of the electron revolving in `3rd` orbit of `M^(5+)` ion

Options

A. `x`
B. `4x`
C. `x//4`
D. `2x`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Kinetic Energy in Bohr's Model
According to Bohr's model, the kinetic energy (KE) of an electron in an orbit is given by the formula:
\[ KE = \frac{Z^2 \cdot e^4 \cdot m}{2 \cdot \hbar^2 \cdot n^2} \]
Where:
- \( Z \) is the atomic number,
- \( e \) is the charge of the electron,
- \( m \) is the mass of the electron,
- \( \hbar \) is the reduced Planck's constant,
- \( n \) is the principal quantum number (orbit number).

### Step 2: Calculate the Kinetic Energy for He\(^+\)
For the He\(^+\) ion:
- \( Z = 2 \) (Helium has an atomic number of 2),
- \( n = 2 \) (we are considering the 2nd orbit).

Using the formula for kinetic energy:
\[ KE_{He^+} = \frac{Z^2}{n^2} = \frac{2^2}{2^2} = 1 \]
Given that this kinetic energy is represented as \( x \), we have:
\[ KE_{He^+} = x \]

### Step 3: Calculate the Ionization Energy for M\(^5+\)
Now, we need to find the ionization energy for an electron in the 3rd orbit of the M\(^5+\) ion:
- Assume \( Z = 6 \) (for M, we assume it is Carbon, which has an atomic number of 6),
- \( n = 3 \).

Using the energy formula for ionization energy:
\[ E = -\frac{Z^2 \cdot 13.6 \, \text{eV}}{n^2} \]
For M\(^5+\):
\[ E_{M^{5+}} = -\frac{6^2 \cdot 13.6}{3^2} = -\frac{36 \cdot 13.6}{9} = -\frac{489.6}{9} = -54.4 \, \text{eV} \]

### Step 4: Relate Ionization Energy to Kinetic Energy
The ionization energy is the energy required to remove an electron from the atom, which is equal to the negative of the total energy of the electron in that orbit. Therefore:
\[ Ionization \, Energy = -E_{M^{5+}} = 54.4 \, \text{eV} \]

### Step 5: Expressing Ionization Energy in terms of x
From Step 2, we know that:
\[ KE_{He^+} = x \]
Thus, we can express the ionization energy of M\(^5+\) in terms of \( x \):
Since \( KE_{He^+} = x \) and the ionization energy for M\(^5+\) can be calculated as:
\[ Ionization \, Energy = 4x \]

### Final Answer
The ionization energy of the electron revolving in the 3rd orbit of M\(^5+\) ion is:
\[ 4x \]

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