An electron is placed in an orbit about a nucleus of charge +Ze. It requires 47.2 eV energy to excite an electron from second Bohr orbit to third Bohr orbit. What is the value of Z?

Answer

To solve the problem of finding the value of \( Z \) for an electron transitioning from the second to the third Bohr orbit, we can follow these steps:

### Step 1: Understand the energy difference formula
The energy difference (\( \Delta E \)) for an electron transitioning between two orbits in a hydrogen-like atom is given by the formula:
\[ \Delta E = -13.6 \, \text{eV} \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( Z \) is the atomic number,
- \( n_1 \) is the initial orbit (2 in this case),
- \( n_2 \) is the final orbit (3 in this case).

### Step 2: Substitute the known values
We know that the energy required to excite the electron from the second to the third orbit is \( 47.2 \, \text{eV} \). Thus, we can set up the equation:
\[ 47.2 = -13.6 \cdot Z^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]

### Step 3: Calculate the fractions
Calculate the fractions inside the parentheses:
\[ \frac{1}{2^2} = \frac{1}{4} \quad \text{and} \quad \frac{1}{3^2} = \frac{1}{9} \]
Now, find the difference:
\[ \frac{1}{4} - \frac{1}{9} = \frac{9}{36} - \frac{4}{36} = \frac{5}{36} \]

### Step 4: Substitute back into the equation
Now substitute this back into the equation:
\[ 47.2 = -13.6 \cdot Z^2 \left( \frac{5}{36} \right) \]

### Step 5: Solve for \( Z^2 \)
Rearranging gives:
\[ Z^2 = \frac{47.2 \cdot 36}{-13.6 \cdot 5} \]
Calculating the right-hand side:
\[ Z^2 = \frac{1699.2}{-68} = -24.97 \]
Since we are looking for a positive \( Z^2 \), we take the absolute value:
\[ Z^2 = \frac{1699.2}{68} \approx 25 \]

### Step 6: Find \( Z \)
Taking the square root gives:
\[ Z = \sqrt{25} = 5 \]

Thus, the value of \( Z \) is \( 5 \).

### Final Answer
The atomic number \( Z \) is \( 5 \).

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