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

Options

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

Answer

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

### Step 2: Calculate the de Broglie wavelength for the electron
For an electron, the momentum \( p \) can be expressed in terms of its kinetic energy \( E \):
\[ p = \sqrt{2mE} \]
where \( m \) is the mass of the electron. Therefore, the de Broglie wavelength for the electron \( \lambda_e \) becomes:
\[ \lambda_e = \frac{h}{p} = \frac{h}{\sqrt{2mE}} \]

### Step 3: Calculate the de Broglie wavelength for the photon
For a photon, the energy \( E \) is related to its momentum \( p \) by the equation:
\[ E = pc \]
where \( c \) is the speed of light. Thus, the momentum of the photon can be expressed as:
\[ p = \frac{E}{c} \]
The de Broglie wavelength for the photon \( \lambda_p \) is:
\[ \lambda_p = \frac{h}{p} = \frac{h}{\frac{E}{c}} = \frac{hc}{E} \]

### Step 4: Find the ratio of the de Broglie wavelengths
Now, we can find the ratio of the de Broglie wavelengths of the electron to the photon:
\[ \frac{\lambda_e}{\lambda_p} = \frac{\frac{h}{\sqrt{2mE}}}{\frac{hc}{E}} = \frac{h}{\sqrt{2mE}} \cdot \frac{E}{hc} \]
This simplifies to:
\[ \frac{\lambda_e}{\lambda_p} = \frac{E}{hc\sqrt{2mE}} = \frac{1}{c} \sqrt{\frac{E}{2m}} \]

### Final Result
Thus, the ratio of the de Broglie wavelengths associated with the electron and the photon is:
\[ \frac{\lambda_e}{\lambda_p} = \frac{1}{c} \sqrt{\frac{E}{2m}} \]

---