An electron revolves round a nucleus of atomic number Z. if 32.4 eV of energy is required to excite an electron from the n=3 state to n=4 state, then the value of Z is

Options

A. 5
B. 6
C. 4
D. 7

Explanation

To solve the problem step by step, we will follow the information provided in the question and apply the relevant physics concepts.

### Step 1: Understand the Energy Levels
The energy of an electron in a hydrogen-like atom (where an electron revolves around a nucleus with atomic number Z) is given by the formula:
\[ E_n = -\frac{13.6 Z^2}{n^2} \text{ eV} \]
where \(E_n\) is the energy of the electron at the principal quantum number \(n\).

### Step 2: Identify the States
We are given that the electron transitions from the \(n=3\) state to the \(n=4\) state. Therefore:
- \(n_1 = 3\)
- \(n_2 = 4\)

### Step 3: Calculate the Energy Difference
The energy difference \(\Delta E\) between these two states can be expressed as:
\[ \Delta E = E_4 - E_3 \]
Substituting the energy formula:
\[ \Delta E = \left(-\frac{13.6 Z^2}{4^2}\right) - \left(-\frac{13.6 Z^2}{3^2}\right) \]
This simplifies to:
\[ \Delta E = -\frac{13.6 Z^2}{16} + \frac{13.6 Z^2}{9} \]

### Step 4: Set the Energy Difference Equal to Given Value
We know from the problem statement that the energy required for this transition is \(32.4 \text{ eV}\):
\[ 32.4 = -\frac{13.6 Z^2}{16} + \frac{13.6 Z^2}{9} \]

### Step 5: Combine the Terms
To combine the terms on the right side, we need a common denominator, which is \(144\):
\[ \Delta E = 32.4 = 13.6 Z^2 \left(-\frac{9}{144} + \frac{16}{144}\right) \]
This simplifies to:
\[ 32.4 = 13.6 Z^2 \left(\frac{7}{144}\right) \]

### Step 6: Solve for \(Z^2\)
Rearranging the equation gives:
\[ Z^2 = \frac{32.4 \times 144}{13.6 \times 7} \]
Calculating the right-hand side:
\[ Z^2 = \frac{4665.6}{95.2} \approx 49 \]
Taking the square root:
\[ Z = 7 \]

### Conclusion
The value of \(Z\) is \(7\).

### Final Answer
Thus, the value of \(Z\) is \(7\).

---