An electron and a photon have same wavelength `lambda`, what is ratio of their kinetic energies ?

Options

A. `h/(m_eclambda)`
B. `(2m_e clambda)/h`
C. `1:1`
D. `h/(2m_eclambda)`

Answer

To find the ratio of the kinetic energies of an electron and a photon that have the same wavelength \( \lambda \), we can follow these steps:

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

### Step 2: Determine the kinetic energy of the electron
The kinetic energy \( K \) of an electron can be expressed in terms of its momentum \( p \):
\[ K_{electron} = \frac{p^2}{2m_e} \]
where \( m_e \) is the mass of the electron.

### Step 3: Relate wavelength to momentum
For both the photon and the electron, we can relate their wavelength \( \lambda \) to momentum. The de Broglie wavelength for the electron is given by:
\[ \lambda = \frac{h}{p} \]
From this, we can express momentum \( p \) in terms of wavelength:
\[ p = \frac{h}{\lambda} \]

### Step 4: Substitute momentum into the kinetic energy equation
Substituting \( p \) into the kinetic energy equation for the electron, we get:
\[ K_{electron} = \frac{(h/\lambda)^2}{2m_e} = \frac{h^2}{2m_e \lambda^2} \]

### Step 5: Write the ratio of the kinetic energies
Now we can find the ratio of the kinetic energies of the electron and the photon:
\[ \text{Ratio} = \frac{K_{electron}}{E_{photon}} = \frac{\frac{h^2}{2m_e \lambda^2}}{\frac{hc}{\lambda}} \]

### Step 6: Simplify the ratio
This simplifies to:
\[ \text{Ratio} = \frac{h^2}{2m_e \lambda^2} \cdot \frac{\lambda}{hc} = \frac{h}{2m_e c \lambda} \]

Thus, the ratio of the kinetic energies of the electron to the photon is:
\[ \text{Ratio} = \frac{h}{2m_e c \lambda} \]