An electron revolves round a nucleus of charge Ze. In order to excite the electron from the `n=20` to `n=3`, the energy required is `47.2 eV`. Z is equal to

Options

A. 3
B. 4
C. 5
D. 2

Explanation

To solve the problem of finding the value of Z when an electron is excited from n=2 to n=3 and the energy required is 47.2 eV, we can follow these steps:

### Step 1: Write down the formula for energy difference
The energy required to excite an electron from one energy level to another in a hydrogen-like atom is given by the formula:
\[ E = 13.6 \, Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( E \) is the energy difference in electron volts (eV),
- \( Z \) is the atomic number (number of protons in the nucleus),
- \( n_1 \) is the initial energy level,
- \( n_2 \) is the final energy level.

### Step 2: Substitute the known values
In this case, we know:
- \( E = 47.2 \, \text{eV} \)
- \( n_1 = 2 \)
- \( n_2 = 3 \)

Substituting these values into the formula gives:
\[ 47.2 = 13.6 \, Z^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]

### Step 3: Calculate the fractions
Calculate \( \frac{1}{2^2} \) and \( \frac{1}{3^2} \):
\[ \frac{1}{2^2} = \frac{1}{4} = 0.25 \]
\[ \frac{1}{3^2} = \frac{1}{9} \approx 0.1111 \]

Now, calculate the difference:
\[ \frac{1}{4} - \frac{1}{9} = 0.25 - 0.1111 = \frac{9 - 4}{36} = \frac{5}{36} \]

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

### Step 5: Simplify the equation
Multiply both sides by \( \frac{36}{5} \):
\[ Z^2 = \frac{47.2 \times 36}{13.6 \times 5} \]

### Step 6: Calculate the right side
Calculating the right side:
\[ Z^2 = \frac{1699.2}{68} \approx 24.98 \]

### Step 7: Take the square root
Taking the square root of both sides gives:
\[ Z \approx \sqrt{25} = 5 \]

### Conclusion
Thus, the value of \( Z \) is:
\[ Z = 5 \]

### Final Answer
The value of \( Z \) is 5.

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