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 (λ_photon) and an electron (λ_electron) for the same energy, we can follow these steps:

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

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

### Step 3: Relate kinetic energy to momentum
The kinetic energy \( E \) of the electron is given by:
\[ E = \frac{1}{2}mv^2 \]
We can rearrange this to express \( v \) in terms of \( E \):
\[ v = \sqrt{\frac{2E}{m}} \]
Now substituting this expression for \( v \) back into the equation for \( p \):
\[ p = m\sqrt{\frac{2E}{m}} = \sqrt{2mE} \]

### Step 4: Substitute momentum back to find λ_e
Now substituting \( p \) back into the expression for \( \lambda_e \):
\[ \lambda_e = \frac{h}{\sqrt{2mE}} \]

### Step 5: Find the wavelength of the photon
For a photon, the energy \( E \) is related to its wavelength \( \lambda \) by:
\[ E = \frac{hc}{\lambda} \]
Rearranging this gives:
\[ \lambda_{photon} = \frac{hc}{E} \]

### Step 6: Find the ratio of the wavelengths
Now we can find the ratio of the wavelengths:
\[ \frac{\lambda_{photon}}{\lambda_e} = \frac{\frac{hc}{E}}{\frac{h}{\sqrt{2mE}}} = \frac{hc \cdot \sqrt{2mE}}{E \cdot h} \]
The \( h \) cancels out, and we simplify:
\[ \frac{\lambda_{photon}}{\lambda_e} = \frac{c \sqrt{2mE}}{E} \]

### Step 7: Final expression
Thus, the final expression for the ratio of the wavelengths is:
\[ \frac{\lambda_{photon}}{\lambda_e} = \frac{c \sqrt{2m}}{E^{1/2}} \]

### Conclusion
The ratio of the wavelengths of the photon to the electron for the same energy is given by:
\[ \frac{\lambda_{photon}}{\lambda_e} = \frac{c \sqrt{2m}}{E^{1/2}} \]