In problems involving electromagenetism it is often convenlent and informative to express answers in terms of a constant, `alpha`, which is a combination of the Coulomb constant, `k_(e) = 1//4piepsi_(0)`, the charge of the electron, e, and `h`, h being Planck's constant. For instant, the lowest energy that a hydrogen atom can have is given by `E = 1//2 alpha^(2) mc^(2)`, wherer m is the mass of the electron and c is the speed of light. Which of the following is the correct expression for `alpha`^2?
(HINt : non-relativistic kinetic energy is `1//2 mv^(2)`, where v is speed.)

Options

A. a.`(k_(e)e^(2))/(hc)`
B. b.`(h)/(k_(e)e^(2)c)`
C. c.`(k_(e)e^(2)h)/(c)`
D. d.`(k_(e)e^(2)pi)/(hc)`

Explanation

To solve the problem, we need to derive the expression for \(\alpha^2\) in terms of the given constants. Let's break down the steps systematically:

### Step 1: Understand the Energies in a Hydrogen Atom
In a hydrogen atom, the total energy \(E\) is the sum of the potential energy \(U\) and the kinetic energy \(K\):
\[ E = U + K \]

### Step 2: Write the Potential Energy
The potential energy \(U\) due to the electrostatic attraction between the proton and the electron is given by:
\[ U = -\frac{e^2}{4\pi \epsilon_0 r} \]

### Step 3: Write the Kinetic Energy
The kinetic energy \(K\) of the electron can be expressed as:
\[ K = \frac{1}{2} mv^2 \]
Since we will relate this to the speed of light \(c\), we can express it as:
\[ K = \frac{1}{2} mc^2 \]

### Step 4: Relate Centripetal Force and Electrostatic Force
For an electron in orbit, the centripetal force is provided by the electrostatic force:
\[ \frac{mv^2}{r} = \frac{e^2}{4\pi \epsilon_0 r^2} \]
Rearranging gives:
\[ mv^2 = \frac{e^2}{4\pi \epsilon_0 r} \]

### Step 5: Substitute for \(mc^2\)
From the centripetal force equation, we can express \(mc^2\) as:
\[ mc^2 = \frac{e^2}{4\pi \epsilon_0 r} \]

### Step 6: Substitute into the Total Energy Equation
Now substitute \(K\) and \(U\) into the total energy equation:
\[ E = -\frac{e^2}{4\pi \epsilon_0 r} + \frac{1}{2} mc^2 \]
Substituting \(mc^2\):
\[ E = -\frac{e^2}{4\pi \epsilon_0 r} + \frac{1}{2} \left(\frac{e^2}{4\pi \epsilon_0 r}\right) \]
This simplifies to:
\[ E = -\frac{1}{2} \frac{e^2}{4\pi \epsilon_0 r} \]

### Step 7: Relate to the Given Expression
According to the problem, the lowest energy of the hydrogen atom is given by:
\[ E = \frac{1}{2} \alpha^2 mc^2 \]
Setting the two expressions for \(E\) equal gives:
\[ -\frac{1}{2} \frac{e^2}{4\pi \epsilon_0 r} = \frac{1}{2} \alpha^2 mc^2 \]
Cancelling \(\frac{1}{2}\) from both sides:
\[ -\frac{e^2}{4\pi \epsilon_0 r} = \alpha^2 mc^2 \]

### Step 8: Solve for \(\alpha^2\)
Now, substituting \(mc^2\) from earlier:
\[ -\frac{e^2}{4\pi \epsilon_0 r} = \alpha^2 \left(\frac{e^2}{4\pi \epsilon_0 r}\right) \]
Thus, we can express \(\alpha^2\):
\[ \alpha^2 = \frac{e^2}{4\pi \epsilon_0 r} \cdot \frac{1}{mc^2} \]

### Step 9: Final Expression
This leads us to the final expression for \(\alpha^2\):
\[ \alpha^2 = \frac{e^2 k_e}{mc^2} \]
where \(k_e = \frac{1}{4\pi \epsilon_0}\).

### Summary
The correct expression for \(\alpha^2\) is:
\[ \alpha^2 = \frac{e^2 k_e}{mc^2} \]