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, we will go through the following steps:

### Step 1: Determine the energy levels for the Bohr model
The energy levels for an electron in a hydrogen-like atom (with nuclear charge +Ze) are given by the formula:
\[ E_n = -\frac{Z^2 \cdot 13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number.

### Step 2: Calculate the energy for the second and third orbits
For the second orbit (\( n=2 \)):
\[ E_2 = -\frac{Z^2 \cdot 13.6}{2^2} = -\frac{Z^2 \cdot 13.6}{4} = -3.4 Z^2 \, \text{eV} \]

For the third orbit (\( n=3 \)):
\[ E_3 = -\frac{Z^2 \cdot 13.6}{3^2} = -\frac{Z^2 \cdot 13.6}{9} \approx -1.51 Z^2 \, \text{eV} \]

### Step 3: Calculate the energy difference between the second and third orbits
The energy required to excite the electron from the second orbit to the third orbit is given as \( 47.2 \, \text{eV} \):
\[ E_3 - E_2 = 47.2 \, \text{eV} \]
Substituting the expressions for \( E_2 \) and \( E_3 \):
\[ \left(-\frac{Z^2 \cdot 13.6}{9}\right) - \left(-\frac{Z^2 \cdot 13.6}{4}\right) = 47.2 \]

### Step 4: Simplify the equation
This simplifies to:
\[ -\frac{Z^2 \cdot 13.6}{9} + \frac{Z^2 \cdot 13.6}{4} = 47.2 \]
Finding a common denominator (36):
\[ -\frac{4Z^2 \cdot 13.6}{36} + \frac{9Z^2 \cdot 13.6}{36} = 47.2 \]
\[ \frac{5Z^2 \cdot 13.6}{36} = 47.2 \]
Multiplying both sides by 36:
\[ 5Z^2 \cdot 13.6 = 47.2 \cdot 36 \]
\[ 5Z^2 \cdot 13.6 = 1699.2 \]
Dividing both sides by 5:
\[ Z^2 \cdot 13.6 = 339.84 \]
Finally, solving for \( Z^2 \):
\[ Z^2 = \frac{339.84}{13.6} \approx 24.99 \]
Taking the square root:
\[ Z \approx 5 \]

### Step 5: Calculate the energy required to excite from the third to the fourth orbit
Using the same energy level formula, we find \( E_4 \):
\[ E_4 = -\frac{Z^2 \cdot 13.6}{4^2} = -\frac{Z^2 \cdot 13.6}{16} = -\frac{5^2 \cdot 13.6}{16} = -\frac{25 \cdot 13.6}{16} = -21.25 \, \text{eV} \]
Now, calculate the energy required to go from \( E_3 \) to \( E_4 \):
\[ E_4 - E_3 = -21.25 - (-1.51 \cdot 25) = -21.25 + 3.4 = -17.85 \, \text{eV} \]
The energy required is:
\[ \Delta E = 17.85 \, \text{eV} \]

### Step 6: Calculate the wavelength of the electromagnetic radiation to remove the electron from the first Bohr orbit to infinity
The energy required to remove the electron from the first orbit to infinity is:
\[ E_1 = -\frac{Z^2 \cdot 13.6}{1^2} = -Z^2 \cdot 13.6 = -25 \cdot 13.6 = -340 \, \text{eV} \]
The energy required to remove it to infinity is:
\[ E = 340 \, \text{eV} \]
Using the relationship between energy and wavelength:
\[ E = \frac{hc}{\lambda} \]
Where \( h = 4.135667696 \times 10^{-15} \, \text{eV s} \) and \( c = 3 \times 10^8 \, \text{m/s} \):
\[ \lambda = \frac{hc}{E} \]
Substituting the values:
\[ \lambda = \frac{(4.135667696 \times 10^{-15} \, \text{eV s})(3 \times 10^8 \, \text{m/s})}{340 \, \text{eV}} \approx 3.65 \times 10^{-7} \, \text{m} = 365 \, \text{nm} \]

### Final Answers
(i) \( Z = 5 \)
(ii) Energy required to excite from the third to the fourth orbit: \( 17.85 \, \text{eV} \)
(iii) Wavelength to remove the electron from the first orbit to infinity: \( 365 \, \text{nm} \)