The orbiting `v_(n) of e^(bar)` in the nth orbit in the case of positronium is x-field compared to that in the nth orbit in a hydrogen atom , where `x` has the value

Options

A. 1
B. `sqrt2`
C. `1// sqrt2`
D. 2

Explanation

To solve the problem of comparing the orbiting velocity \( v_n \) of the electron in the nth orbit of positronium with that in the nth orbit of a hydrogen atom, we can follow these steps:

### Step 1: Understand the System
In a hydrogen atom, we have a single electron orbiting a proton, while in positronium, we have an electron and a positron (the electron's antiparticle) orbiting around their common center of mass. The mass of the positron is equal to the mass of the electron.

### Step 2: Write the Centripetal Force Equation
For both systems, the centripetal force required to keep the electron in orbit can be expressed as:
\[ \frac{mv^2}{r} = F \]
where \( m \) is the mass of the particle, \( v \) is the orbital velocity, and \( r \) is the radius of the orbit.

### Step 3: Analyze the Hydrogen Atom
For the hydrogen atom, the force acting on the electron is due to the electrostatic attraction between the electron and the proton:
\[ F = \frac{k \cdot e^2}{r^2} \]
where \( k \) is Coulomb's constant and \( e \) is the charge of the electron. Setting the centripetal force equal to the electrostatic force gives:
\[ \frac{mv^2}{r} = \frac{k \cdot e^2}{r^2} \]
Rearranging this, we find:
\[ mv^2 = \frac{k \cdot e^2}{r} \]

### Step 4: Analyze the Positronium
For positronium, the effective mass of the system is different. The reduced mass \( \mu \) of the electron-positron system is given by:
\[ \mu = \frac{m_e \cdot m_e}{m_e + m_e} = \frac{m_e}{2} \]
The radius of the orbit in positronium is also different. The radius \( r_{ps} \) for positronium is approximately twice that of hydrogen due to the presence of both an electron and a positron:
\[ r_{ps} = 2r \]
Now, we can write the centripetal force equation for positronium:
\[ \frac{\mu v_{ps}^2}{r_{ps}} = \frac{k \cdot e^2}{r_{ps}^2} \]
Substituting \( \mu \) and \( r_{ps} \):
\[ \frac{\frac{m_e}{2} v_{ps}^2}{2r} = \frac{k \cdot e^2}{(2r)^2} \]
This simplifies to:
\[ \frac{m_e v_{ps}^2}{4r} = \frac{k \cdot e^2}{4r^2} \]
Thus, we have:
\[ m_e v_{ps}^2 = \frac{k \cdot e^2}{r} \]

### Step 5: Compare the Two Systems
From the equations derived for hydrogen and positronium:
1. For hydrogen: \( mv^2 = \frac{k \cdot e^2}{r} \)
2. For positronium: \( m_e v_{ps}^2 = \frac{k \cdot e^2}{r} \)

Since both equations are equal, we find that:
\[ v_{ps} = v \]

### Conclusion
Thus, the orbiting velocity \( v_n \) of the electron in the nth orbit of positronium is the same as that in the nth orbit of hydrogen. Therefore, the value of \( x \) is:
\[ \boxed{1} \]