An electron of mass `m` and a photon have same energy `E`. The ratio of de - Broglie wavelengths associated with them is :

Options

A. `(1)/( C) ((E )/( 2 m))^(1//2)`
B. `((E )/(2 m))^(1//2)`
C. `C ( 2m E)^(1//2)`
D. `(1)/(xC) ((2 m)/( E))^(1//2)`

Explanation

To find the ratio of the de Broglie wavelengths associated with an electron and a photon that have the same energy \( E \), we can follow these steps:

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

- The energy of the electron can be expressed as its kinetic energy:
\[ E = \frac{1}{2}mv^2 \]
where \( m \) is the mass of the electron and \( v \) is its velocity.

### Step 2: Find the de Broglie wavelength of the electron
- The de Broglie wavelength \( \lambda_E \) of the electron is given by:
\[ \lambda_E = \frac{h}{p} \]
where \( p \) is the momentum of the electron. The momentum \( p \) can be expressed in terms of kinetic energy:
\[ p = mv = \sqrt{2mE} \]
Therefore, the de Broglie wavelength of the electron becomes:
\[ \lambda_E = \frac{h}{\sqrt{2mE}} \]

### Step 3: Find the de Broglie wavelength of the photon
- For the photon, the momentum \( p \) is related to its energy by:
\[ p = \frac{E}{c} \]
Thus, the de Broglie wavelength of the photon is:
\[ \lambda_P = \frac{h}{p} = \frac{hc}{E} \]

### Step 4: Calculate the ratio of the de Broglie wavelengths
- Now, we can find the ratio of the de Broglie wavelengths:
\[ \frac{\lambda_E}{\lambda_P} = \frac{\frac{h}{\sqrt{2mE}}}{\frac{hc}{E}} = \frac{h}{\sqrt{2mE}} \cdot \frac{E}{hc} \]
Simplifying this gives:
\[ \frac{\lambda_E}{\lambda_P} = \frac{E}{hc\sqrt{2mE}} = \frac{1}{c\sqrt{2m}} \sqrt{E} \]

### Conclusion
- The final ratio of the de Broglie wavelengths associated with the electron and the photon is:
\[ \frac{\lambda_E}{\lambda_P} = \frac{1}{c\sqrt{2m}} \sqrt{E} \]