The ionization energy of H atom is x kJ. The energy required for the electron to jump from `n = 2` to `n = 3` will be:

Options

A. 5x
B. `36x//5`
C. `5x//36`
D. 9x/4

Answer

To solve the problem of finding the energy required for an electron to jump from \( n = 2 \) to \( n = 3 \) in a hydrogen atom, we can follow these steps:

### Step 1: Understand Ionization Energy
The ionization energy of a hydrogen atom is the energy required to remove the electron completely from the atom, which occurs when the electron transitions from the ground state (\( n = 1 \)) to a state where \( n \) approaches infinity.

### Step 2: Use the Ionization Energy Formula
The formula for the ionization energy (\( E \)) of a hydrogen atom is given by:
\[ E = -\frac{13.6 \, \text{eV}}{n^2} \]
For ionization from \( n = 1 \) to \( n = \infty \):
\[ E = 13.6 \, \text{eV} \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = 13.6 \, \text{eV} \left( 1 - 0 \right) = 13.6 \, \text{eV} \]

### Step 3: Relate Ionization Energy to Given Value
According to the problem, the ionization energy is given as \( x \, \text{kJ} \). We know from our calculation that:
\[ x = 13.6 \, \text{eV} \]

### Step 4: Calculate Energy for Transition from \( n = 2 \) to \( n = 3 \)
The energy required for the electron to jump from \( n = 2 \) to \( n = 3 \) can be calculated using the same formula:
\[ E = 13.6 \, \text{eV} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Substituting \( n_1 = 2 \) and \( n_2 = 3 \):
\[ E = 13.6 \, \text{eV} \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 13.6 \, \text{eV} \left( \frac{1}{4} - \frac{1}{9} \right) \]

### Step 5: Simplify the Expression
Now, calculate \( \frac{1}{4} - \frac{1}{9} \):
\[ \frac{1}{4} = \frac{9}{36}, \quad \frac{1}{9} = \frac{4}{36} \]
Thus,
\[ \frac{1}{4} - \frac{1}{9} = \frac{9}{36} - \frac{4}{36} = \frac{5}{36} \]

### Step 6: Final Calculation
Now substitute back into the energy equation:
\[ E = 13.6 \, \text{eV} \cdot \frac{5}{36} = \frac{68}{36} \, \text{eV} = \frac{5 \cdot 13.6}{36} \, \text{eV} \]
Since \( x = 13.6 \, \text{eV} \), we can express the energy as:
\[ E = \frac{5x}{36} \, \text{eV} \]

### Conclusion
The energy required for the electron to jump from \( n = 2 \) to \( n = 3 \) is:
\[ \frac{5x}{36} \, \text{eV} \]