An electron and a photon possess the same de Broglie wavelength. If `E_e` and `E_ph` are, respectively, the energies of electron and photon while v and c are their respective velocities, then `(E_e)/(E_(ph))` is equal to

Options

A. `(v)/(c )`
B. `(v)/(2c)`
C. `(v)/(3c)`
D. `(v)/(4c)`

Explanation

To solve the problem, we need to find the ratio of the energies of an electron and a photon that possess the same de Broglie wavelength. Let's denote the energies of the electron and photon as \(E_e\) and \(E_{ph}\), respectively, and their velocities as \(v\) and \(c\) (where \(c\) is the speed of light).

### Step 1: Write the de Broglie wavelength equations
The de Broglie wavelength \(\lambda\) for an electron is given by:
\[ \lambda = \frac{h}{mv} \]
where \(h\) is Planck's constant, \(m\) is the mass of the electron, and \(v\) is its velocity.

For a photon, the de Broglie wavelength is given by:
\[ \lambda = \frac{h}{E_{ph}/c} \]
where \(E_{ph}\) is the energy of the photon and \(c\) is the speed of light.

### Step 2: Set the two expressions for wavelength equal
Since both the electron and photon have the same de Broglie wavelength, we can set the two equations equal to each other:
\[ \frac{h}{mv} = \frac{h}{E_{ph}/c} \]

### Step 3: Simplify the equation
By canceling \(h\) from both sides, we get:
\[ \frac{1}{mv} = \frac{c}{E_{ph}} \]
Cross-multiplying gives us:
\[ E_{ph} = mvc \]

### Step 4: Express the energy of the electron
The energy of the electron can be expressed in terms of its kinetic energy:
\[ E_e = \frac{1}{2}mv^2 \]

### Step 5: Find the ratio \(\frac{E_e}{E_{ph}}\)
Now, we can find the ratio of the energies:
\[ \frac{E_e}{E_{ph}} = \frac{\frac{1}{2}mv^2}{mvc} \]
This simplifies to:
\[ \frac{E_e}{E_{ph}} = \frac{v}{2c} \]

### Conclusion
Thus, the final expression for the ratio of the energies of the electron and photon is:
\[ \frac{E_e}{E_{ph}} = \frac{v}{2c} \]