A hydrogen-like atom has one electron revolving around a stationary nucleus. The energy required to excite the electron from the second orbit to the third orbit is 47.2 eV. The atomic number of the atom is

Options

A. 3
B. 4
C. 5
D. 6

Explanation

To find the atomic number (Z) of a hydrogen-like atom given the energy required to excite the electron from the second orbit to the third orbit, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Energy Formula**:
The energy of an electron in a hydrogen-like atom 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 in the nth orbit, and Z is the atomic number.

2. **Calculate the Energy Difference**:
The energy required to excite the electron from the second orbit (n1 = 2) to the third orbit (n2 = 3) is given by:
\[ \Delta E = E_3 - E_2 \]
Substituting the energy values:
\[ \Delta E = \left(-\frac{13.6 \, Z^2}{3^2}\right) - \left(-\frac{13.6 \, Z^2}{2^2}\right) \]
\[ \Delta E = -\frac{13.6 \, Z^2}{9} + \frac{13.6 \, Z^2}{4} \]

3. **Finding a Common Denominator**:
The common denominator for 9 and 4 is 36. Thus, we rewrite the equation:
\[ \Delta E = 13.6 \, Z^2 \left(-\frac{4}{36} + \frac{9}{36}\right) \]
\[ \Delta E = 13.6 \, Z^2 \left(\frac{5}{36}\right) \]

4. **Set the Energy Difference Equal to Given Value**:
We know from the problem that \(\Delta E = 47.2 \, \text{eV}\):
\[ 47.2 = 13.6 \, Z^2 \left(\frac{5}{36}\right) \]

5. **Solve for Z**:
Rearranging the equation gives:
\[ Z^2 = \frac{47.2 \times 36}{13.6 \times 5} \]
Calculating the right-hand side:
\[ Z^2 = \frac{1699.2}{68} = 24.97 \approx 25 \]
Therefore, taking the square root:
\[ Z = 5 \]

### Conclusion:
The atomic number of the atom is \(Z = 5\).