Consider a system containing a negatively charge poin `(pi, m_(pi) = 273^(@)m_(e))` orbital around a staionary nucleus of atomic number Z .The total energy `(E_(n))` of ion is half of its potential energy `(PE_(n))` in nth sationary state .The motion of the poin can be assumed to be in a uniform circular notion with centripents force given by the force of attaraction between the positive uncless and the point .Assume that point revolves only in the stationary satte defined by the quantisation of its angular momentum about the nucless as Bohr's model
The longest wavelength radiation emitted in the emission spectrum when the pion de-excited from `n = 3` to ground state lies which of the following region ?

Options

A. UV
B. Visible
C. Intire-Red
D. Cannot be calculated

Explanation

To solve the problem, we need to determine the longest wavelength radiation emitted when a pion transitions from the n = 3 state to the ground state (n = 1) in a hydrogen-like atom with atomic number Z.

### Step-by-step Solution:

1. **Understanding the Energy Levels**:
In a hydrogen-like atom, the energy levels can be expressed as:
\[ E_n = -\frac{Z^2 \cdot 13.6 \, \text{eV}}{n^2} \]
where \( Z \) is the atomic number and \( n \) is the principal quantum number.

2. **Calculating the Energy for n = 3 and n = 1**:
- For \( n = 3 \):
\[ E_3 = -\frac{Z^2 \cdot 13.6}{3^2} = -\frac{Z^2 \cdot 13.6}{9} \, \text{eV} \]
- For \( n = 1 \):
\[ E_1 = -Z^2 \cdot 13.6 \, \text{eV} \]

3. **Finding the Energy Difference**:
The energy difference \( \Delta E \) when the pion transitions from \( n = 3 \) to \( n = 1 \) is:
\[ \Delta E = E_1 - E_3 = \left(-Z^2 \cdot 13.6\right) - \left(-\frac{Z^2 \cdot 13.6}{9}\right) \]
Simplifying this gives:
\[ \Delta E = -Z^2 \cdot 13.6 + \frac{Z^2 \cdot 13.6}{9} = Z^2 \cdot 13.6 \left(1 - \frac{1}{9}\right) = Z^2 \cdot 13.6 \cdot \frac{8}{9} \]

4. **Calculating the Wavelength**:
Using the relationship between energy and wavelength:
\[ E = \frac{hc}{\lambda} \]
Rearranging gives:
\[ \lambda = \frac{hc}{\Delta E} \]
Substituting \( \Delta E \):
\[ \lambda = \frac{hc}{Z^2 \cdot 13.6 \cdot \frac{8}{9}} \]

5. **Identifying the Region of Wavelength**:
Since \( \lambda \) depends on \( Z \), we cannot calculate a specific numerical value without knowing \( Z \). However, we can analyze the wavelength's dependence on \( Z \):
- For hydrogen (\( Z = 1 \)), the wavelength will be in the ultraviolet region.
- For higher \( Z \), the wavelength decreases, moving into the X-ray region.

### Conclusion:
Since the question does not specify the atomic number \( Z \), we cannot determine an exact wavelength or the region it falls into. However, we can conclude that the longest wavelength emitted during this transition will generally lie in the ultraviolet region for hydrogen-like atoms.