For same energy , find the ratio of ` lambda _("photon" ) and lambda _("electron") ` (Here m is mass of electron)

Options

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

Answer

To find the ratio of the wavelengths of a photon and an electron for the same energy, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Energy Equivalence**:
Given that the energy \( E \) is the same for both the photon and the electron, we can denote this energy as \( E \).

2. **Photon Momentum**:
The momentum \( p \) of a photon can be expressed using the relationship:
\[ p_{\text{photon}} = \frac{E}{c} \]
where \( c \) is the speed of light.

3. **Electron Momentum**:
The momentum \( p \) of an electron can be expressed using the kinetic energy formula:
\[ p_{\text{electron}} = \sqrt{2mE} \]
where \( m \) is the mass of the electron.

4. **De Broglie Wavelength**:
According to de Broglie's hypothesis, the wavelength \( \lambda \) is inversely proportional to momentum:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant.

5. **Wavelength of Photon**:
The wavelength of the photon can be expressed as:
\[ \lambda_{\text{photon}} = \frac{h}{p_{\text{photon}}} = \frac{h}{\frac{E}{c}} = \frac{hc}{E} \]

6. **Wavelength of Electron**:
The wavelength of the electron can be expressed as:
\[ \lambda_{\text{electron}} = \frac{h}{p_{\text{electron}}} = \frac{h}{\sqrt{2mE}} \]

7. **Finding the Ratio**:
Now, we can find the ratio of the wavelengths:
\[ \frac{\lambda_{\text{photon}}}{\lambda_{\text{electron}}} = \frac{\frac{hc}{E}}{\frac{h}{\sqrt{2mE}}} \]
Simplifying this gives:
\[ \frac{\lambda_{\text{photon}}}{\lambda_{\text{electron}}} = \frac{hc \cdot \sqrt{2mE}}{E \cdot h} = \frac{c \cdot \sqrt{2m}}{E^{1/2}} \]

8. **Final Expression**:
Thus, the ratio of the wavelengths is:
\[ \frac{\lambda_{\text{photon}}}{\lambda_{\text{electron}}} = \frac{c \sqrt{2m}}{\sqrt{E}} \]

### Final Result:
The ratio of the wavelength of the photon to the wavelength of the electron for the same energy is:
\[ \frac{\lambda_{\text{photon}}}{\lambda_{\text{electron}}} = \frac{c \sqrt{2m}}{\sqrt{E}} \]