An electron & a photon have same energy E. Find the ratio of de Broglie wavelength of electron to wavelength of photon. Given mass of electron is m & speed of light is C

Options

A. `1/C sqrt(E/(2m))`
B. `C sqrt(E/m)`
C. `1/C sqrt(2mE)`
D. `1/C sqrt(E/m)`

Explanation

To find the ratio of the de Broglie wavelength of an electron to the wavelength of a photon, we can follow these steps:

### Step 1: Understand the Energy of the Photon
The energy \( E \) of a photon is given by the equation:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the photon.

### Step 2: Solve for the Wavelength of the Photon
From the equation above, we can rearrange it to solve for the wavelength \( \lambda \) of the photon:
\[ \lambda_{\text{photon}} = \frac{hc}{E} \]

### Step 3: Understand the Energy of the Electron
The energy \( E \) of an electron can be expressed in terms of its mass \( m \) and velocity \( v \) using the kinetic energy formula:
\[ E = \frac{1}{2}mv^2 \]

### Step 4: Find the de Broglie Wavelength of the Electron
The de Broglie wavelength \( \lambda \) of a particle is given by:
\[ \lambda_{\text{electron}} = \frac{h}{p} \]
where \( p \) is the momentum of the electron. The momentum \( p \) can be expressed as:
\[ p = mv \]
Thus, the de Broglie wavelength of the electron becomes:
\[ \lambda_{\text{electron}} = \frac{h}{mv} \]

### Step 5: Relate the Energy to the Velocity of the Electron
From the kinetic energy equation, we can express the velocity \( v \) in terms of energy \( E \):
\[ E = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{2E}{m}} \]

### Step 6: Substitute the Velocity into the de Broglie Wavelength
Now, substituting \( v \) back into the equation for the de Broglie wavelength of the electron:
\[ \lambda_{\text{electron}} = \frac{h}{m\sqrt{\frac{2E}{m}}} = \frac{h}{\sqrt{2mE}} \]

### Step 7: Find the Ratio of the Wavelengths
Now we can find the ratio of the de Broglie wavelength of the electron to the wavelength of the photon:
\[ \frac{\lambda_{\text{electron}}}{\lambda_{\text{photon}}} = \frac{\frac{h}{\sqrt{2mE}}}{\frac{hc}{E}} = \frac{hE}{hc\sqrt{2mE}} = \frac{E}{c\sqrt{2mE}} \]

### Step 8: Simplify the Ratio
This simplifies to:
\[ \frac{\lambda_{\text{electron}}}{\lambda_{\text{photon}}} = \frac{1}{c} \sqrt{\frac{E}{2m}} \]

### Final Answer
Thus, the ratio of the de Broglie wavelength of the electron to the wavelength of the photon is:
\[ \frac{\lambda_{\text{electron}}}{\lambda_{\text{photon}}} = \frac{1}{c} \sqrt{\frac{E}{2m}} \]