Consider a system containing a negatively charge point `(pi, m_(pi) = 273m_(e))` orbital around a stationary nucleus of atomic number Z .The total energy `(E_(n))` of ion is half of its potential energy `(PE_(n))` in nth stationary state .The motion of the point can be assumed to be in a uniform circular motion with centripetal force given by the force of attraction between the positive nucleus and the point .Assume that point revolves only in the stationary state defined by the quantization of its angular momentum about the nucleus as Bohr's model
Number of waves made by the point when it orbits in third excitation state are

Options

A. `3`
B. `4`
C. `3Z^(2)`
D. `4Z^(2)`

Explanation

To solve the problem step by step, we will follow the concepts of Bohr's model and de Broglie's hypothesis.

### Step 1: Understand the System
We have a negatively charged particle (let's call it Pi) with mass \( m_{\pi} = 273 m_e \) (where \( m_e \) is the mass of an electron) orbiting around a nucleus with atomic number \( Z \). The total energy \( E_n \) of the ion is half of its potential energy \( PE_n \) in the nth stationary state.

### Step 2: Relate Total Energy and Potential Energy
According to the problem, we have:
\[ E_n = \frac{1}{2} PE_n \]
In the context of the Bohr model, the total energy \( E_n \) is given by:
\[ E_n = -\frac{Z^2 e^4 m_{\pi}}{2 \hbar^2 n^2} \]
And the potential energy \( PE_n \) is:
\[ PE_n = -\frac{Z e^2}{r} \]
From the relationship given, we can deduce that:
\[ PE_n = -2E_n \]

### Step 3: Identify the Excited State
The problem states that we are looking for the number of waves made by the point when it orbits in the third excited state. The third excited state corresponds to \( n = 4 \) (since the ground state is \( n = 1 \)).

### Step 4: Apply de Broglie's Hypothesis
According to de Broglie's hypothesis, the wavelength \( \lambda \) associated with the particle is given by:
\[ \lambda = \frac{h}{m_{\pi} v} \]
where \( v \) is the velocity of the particle.

### Step 5: Quantization of Angular Momentum
From Bohr's model, the quantization of angular momentum states that:
\[ m_{\pi} v r = n \frac{h}{2\pi} \]
For \( n = 4 \):
\[ m_{\pi} v r = 4 \frac{h}{2\pi} \]

### Step 6: Relate Wavelength to Radius
From the relationship derived from the de Broglie wavelength and the quantization of angular momentum, we can write:
\[ 2\pi r = n \lambda \]
Substituting \( n = 4 \):
\[ 2\pi r = 4 \lambda \]

### Step 7: Calculate the Number of Waves
The number of waves made by the point in the orbit is equal to \( n \):
\[ \text{Number of waves} = n = 4 \]

### Final Answer
The number of waves made by the point when it orbits in the third excitation state is **4**.

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