Supose the potential energy between electron and porton a distance r is given by `U=-(Ke^(2))/(3r^(3))` Assuming Bohar's Model to be valid for this atom if the speed of elctron in `n^(th)` orbit depends on the prinicpal quantum number n as v `oon^(x)` then find the vlaue of x.

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to analyze the potential energy \( U \) given by:

\[ U = -\frac{K e^2}{3r^3} \]

where \( K \) is a constant, \( e \) is the charge of the electron, and \( r \) is the distance between the electron and the proton. We will use the Bohr model to find the relationship between the speed of the electron in the \( n^{th} \) orbit and the principal quantum number \( n \).

### Step 1: Calculate the Force
The force \( F \) between the electron and proton can be found by taking the negative gradient of the potential energy:

\[ F = -\frac{dU}{dr} \]

Differentiating \( U \):

\[ \frac{dU}{dr} = -\frac{d}{dr} \left(\frac{K e^2}{3r^3}\right) = -\left(-\frac{3K e^2}{3r^4}\right) = \frac{K e^2}{r^4} \]

Thus, the force \( F \) is:

\[ F = \frac{K e^2}{r^4} \]

### Step 2: Relate Force to Centripetal Force
According to the Bohr model, the centripetal force required to keep the electron in circular motion is given by:

\[ F = \frac{mv^2}{r} \]

where \( m \) is the mass of the electron and \( v \) is its speed. Setting the two expressions for force equal gives:

\[ \frac{K e^2}{r^4} = \frac{mv^2}{r} \]

### Step 3: Rearranging the Equation
Rearranging the equation to solve for \( v^2 \):

\[ mv^2 = \frac{K e^2}{r^3} \]

Thus,

\[ v^2 = \frac{K e^2}{mr^3} \]

### Step 4: Expressing \( r \) in terms of \( n \)
From the Bohr model, we know that the radius \( r \) of the \( n^{th} \) orbit is proportional to \( n^2 \):

\[ r \propto n^2 \]

This implies:

\[ r^3 \propto n^6 \]

### Step 5: Substitute \( r^3 \) into the Velocity Equation
Substituting \( r^3 \) back into the equation for \( v^2 \):

\[ v^2 \propto \frac{K e^2}{m n^6} \]

Taking the square root to find \( v \):

\[ v \propto \frac{1}{n^3} \]

### Step 6: Conclusion
Thus, we can express the speed of the electron in the \( n^{th} \) orbit as:

\[ v \propto n^{-3} \]

This means that \( v \) is proportional to \( n^{-3} \), and therefore, the value of \( x \) is:

\[ x = -3 \]

### Final Answer
The value of \( x \) is \( -3 \).

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