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`

Explanation

To find 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 given as \( x \) kJ. Ionization energy is the energy required to remove the most loosely bound electron from an atom in its gaseous state. For hydrogen, this corresponds to the transition from \( n = 1 \) to \( n = \infty \).

### Step 2: Energy Levels in Hydrogen
The energy of an electron in a hydrogen atom at a given principal quantum number \( n \) is given by the formula:
\[ E_n = -\frac{k}{n^2} \]
where \( k \) is a constant. For hydrogen, the ionization energy can be expressed as:
\[ E_{\infty} - E_1 = x \]
where \( E_{\infty} = 0 \) (energy at infinity) and \( E_1 = -k \).

### Step 3: Relate Ionization Energy to Constant \( k \)
From the ionization energy expression:
\[ 0 - (-k) = x \implies k = x \]

### Step 4: Calculate Energy for Transition from \( n = 2 \) to \( n = 3 \)
We need to find the energy difference between the states \( n = 3 \) and \( n = 2 \):
\[ E_2 = -\frac{k}{2^2} = -\frac{k}{4} \]
\[ E_3 = -\frac{k}{3^2} = -\frac{k}{9} \]
Now, calculate the energy required for the transition:
\[ \Delta E = E_3 - E_2 = \left(-\frac{k}{9}\right) - \left(-\frac{k}{4}\right) \]
\[ \Delta E = -\frac{k}{9} + \frac{k}{4} \]
To combine these fractions, find a common denominator (36):
\[ \Delta E = -\frac{4k}{36} + \frac{9k}{36} = \frac{5k}{36} \]

### Step 5: Substitute \( k \) with \( x \)
Since we found that \( k = x \):
\[ \Delta E = \frac{5x}{36} \]

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

---