A single electron orbits around a stationary nucleus of charge `+Ze`, where Z is a constant and e is the electronic charge. It requires `47.2 eV` to excited the electron from the second Bohr orbit to the third Bohr orbit. Find (i) the value of Z, (ii) the energy required to excite the electron from the third to the fourth Bohr orbit and (iii) the wavelength of the electromagnetic radiation radiation to remove the electron from the first Bohr orbit to infinity.

Answer

To solve the problem step by step, we will break it down into three parts as mentioned in the question.

### Step 1: Finding the value of Z

1. **Understanding the energy difference**: The energy required to excite the electron from the second orbit (n=2) to the third orbit (n=3) is given as 47.2 eV. According to the Bohr model, the energy of an electron in the nth orbit is given by:
\[ E_n = -\frac{Z^2 \cdot 13.6 \, \text{eV}}{n^2} \]
Therefore, the energy difference between the third and second orbits can be expressed as:
\[ E_3 - E_2 = -\frac{Z^2 \cdot 13.6}{3^2} - \left(-\frac{Z^2 \cdot 13.6}{2^2}\right) \]
Simplifying this gives:
\[ E_3 - E_2 = Z^2 \cdot 13.6 \left(\frac{1}{2^2} - \frac{1}{3^2}\right) \]

2. **Calculating the energy difference**:
\[ E_3 - E_2 = Z^2 \cdot 13.6 \left(\frac{1}{4} - \frac{1}{9}\right) = Z^2 \cdot 13.6 \left(\frac{9 - 4}{36}\right) = Z^2 \cdot 13.6 \cdot \frac{5}{36} \]
Setting this equal to 47.2 eV:
\[ Z^2 \cdot 13.6 \cdot \frac{5}{36} = 47.2 \]

3. **Solving for Z**:
\[ Z^2 = \frac{47.2 \cdot 36}{13.6 \cdot 5} \]
\[ Z^2 = \frac{1699.2}{68} = 24.97 \]
Taking the square root:
\[ Z \approx 5 \]

### Step 2: Finding the energy required to excite the electron from the third to the fourth orbit

1. **Energy difference from n=3 to n=4**:
\[ E_4 - E_3 = Z^2 \cdot 13.6 \left(\frac{1}{3^2} - \frac{1}{4^2}\right) \]
\[ E_4 - E_3 = Z^2 \cdot 13.6 \left(\frac{1}{9} - \frac{1}{16}\right) \]

2. **Calculating the energy difference**:
\[ E_4 - E_3 = Z^2 \cdot 13.6 \left(\frac{16 - 9}{144}\right) = Z^2 \cdot 13.6 \cdot \frac{7}{144} \]
Substituting \(Z = 5\):
\[ E_4 - E_3 = 5^2 \cdot 13.6 \cdot \frac{7}{144} = 25 \cdot 13.6 \cdot \frac{7}{144} \]
\[ E_4 - E_3 = \frac{25 \cdot 13.6 \cdot 7}{144} \approx 16.53 \, \text{eV} \]

### Step 3: Finding the wavelength of the electromagnetic radiation to remove the electron from the first Bohr orbit to infinity

1. **Energy from n=1 to infinity**:
\[ E_{\infty} - E_1 = Z^2 \cdot 13.6 \left(\frac{1}{1^2} - \frac{1}{\infty^2}\right) = Z^2 \cdot 13.6 \]
Substituting \(Z = 5\):
\[ E_{\infty} - E_1 = 5^2 \cdot 13.6 = 25 \cdot 13.6 = 340 \, \text{eV} \]

2. **Finding the wavelength**:
Using the relation \(E = \frac{hc}{\lambda}\):
\[ \lambda = \frac{hc}{E} \]
Where \(h = 6.63 \times 10^{-34} \, \text{Js}\) and \(c = 3 \times 10^8 \, \text{m/s}\):
\[ \lambda = \frac{6.63 \times 10^{-34} \cdot 3 \times 10^8}{340 \cdot 1.6 \times 10^{-19}} \]
Converting eV to Joules (1 eV = \(1.6 \times 10^{-19}\) J):
\[ \lambda \approx 36.5 \, \text{Angstroms} \]

### Summary of Results:
(i) \(Z = 5\)
(ii) Energy required from n=3 to n=4 is approximately \(16.53 \, \text{eV}\)
(iii) Wavelength to remove the electron from n=1 to infinity is approximately \(36.5 \, \text{Angstroms}\)