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 have the same de Broglie wavelength. Let's break it down step by step.

### Step 1: Write the de Broglie wavelength formula
The de Broglie wavelength (λ) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum.

### Step 2: Find the de Broglie wavelength for the electron
For an electron, the momentum \( p \) can be expressed in terms of its mass \( m \) and velocity \( v \):
\[ p_e = mv \]
Thus, the de Broglie wavelength for the electron is:
\[ \lambda_e = \frac{h}{mv} \]

### Step 3: Find the de Broglie wavelength for the photon
For a photon, the momentum is related to its energy \( E_{ph} \) by the equation:
\[ p_{ph} = \frac{E_{ph}}{c} \]
where \( c \) is the speed of light. The de Broglie wavelength for the photon can be expressed as:
\[ \lambda_{ph} = \frac{h}{p_{ph}} = \frac{h}{\frac{E_{ph}}{c}} = \frac{hc}{E_{ph}} \]

### Step 4: Set the wavelengths equal
Since the problem states that the electron and photon possess the same de Broglie wavelength, we can set the two equations equal to each other:
\[ \frac{h}{mv} = \frac{hc}{E_{ph}} \]

### Step 5: Cancel \( h \) from both sides
By canceling \( h \) from both sides, we get:
\[ \frac{1}{mv} = \frac{c}{E_{ph}} \]

### Step 6: Rearranging the equation
Rearranging the equation gives:
\[ E_{ph} = mvc \]

### Step 7: Express the energy of the electron
The energy of the electron \( E_e \) can be expressed as its kinetic energy:
\[ E_e = \frac{1}{2}mv^2 \]

### Step 8: Find the ratio of energies
Now, we can find the ratio \( \frac{E_e}{E_{ph}} \):
\[ \frac{E_e}{E_{ph}} = \frac{\frac{1}{2}mv^2}{mvc} \]
This simplifies to:
\[ \frac{E_e}{E_{ph}} = \frac{v}{2c} \]

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