If the difference of energies of an electron in the second and the fourth orbits of an atom is E. Find the ionisation energy of that atom.

Options

A. `(12 E)/(3)`
B. `(8E)/(2)`
C. `(16E)/(3)`
D. `(15E)/(2)`.

Answer

To solve the problem, we need to find the ionization energy of an atom given the energy difference between the second and fourth orbits. Let's break down the steps:

### Step-by-Step Solution:

1. **Understanding the Energy Levels**:
According to Bohr's model, the energy of an electron in the nth orbit of an atom is given by the formula:
\[ E_n = -\frac{13.6 Z^2}{n^2} \text{ eV} \]
where \( Z \) is the atomic number and \( n \) is the orbit number.

2. **Calculate the Energy for the 4th Orbit**:
For the 4th orbit (\( n = 4 \)):
\[ E_4 = -\frac{13.6 Z^2}{4^2} = -\frac{13.6 Z^2}{16} \]

3. **Calculate the Energy for the 2nd Orbit**:
For the 2nd orbit (\( n = 2 \)):
\[ E_2 = -\frac{13.6 Z^2}{2^2} = -\frac{13.6 Z^2}{4} \]

4. **Find the Energy Difference**:
The difference in energy between the second and fourth orbits is given as \( E \):
\[ E = E_2 - E_4 \]
Substituting the expressions for \( E_2 \) and \( E_4 \):
\[ E = \left(-\frac{13.6 Z^2}{4}\right) - \left(-\frac{13.6 Z^2}{16}\right) \]
Simplifying this:
\[ E = -\frac{13.6 Z^2}{4} + \frac{13.6 Z^2}{16} \]
To combine these fractions, we find a common denominator (16):
\[ E = -\frac{13.6 Z^2 \cdot 4}{16} + \frac{13.6 Z^2}{16} = \frac{-54.4 Z^2 + 13.6 Z^2}{16} \]
\[ E = \frac{-40.8 Z^2}{16} = -\frac{2.55 Z^2}{1} \text{ eV} \]

5. **Rearranging for Ionization Energy**:
The ionization energy (IE) is the energy required to remove an electron from the ground state (n=1) to infinity. The energy at the first orbit is:
\[ E_1 = -\frac{13.6 Z^2}{1^2} = -13.6 Z^2 \]
The energy at infinity is 0. Thus, the ionization energy is:
\[ \text{IE} = E_{\infty} - E_1 = 0 - (-13.6 Z^2) = 13.6 Z^2 \]

6. **Expressing Ionization Energy in Terms of E**:
From our earlier calculation, we found:
\[ 13.6 Z^2 = \frac{16}{3} E \]
Therefore, the ionization energy can be expressed as:
\[ \text{IE} = \frac{16}{3} E \]

### Final Answer:
The ionization energy of the atom is:
\[ \text{Ionization Energy} = \frac{16}{3} E \]