A photon and an electron moving with a velocity v have the same de broglie wavelength . Then the ratio of the kinetic energy of the electron to the kinetic energy of the photon is [C is the speed of light ]

Options

A. a) `v/C`
B. b) `(2v)/(C )`
C. c) `(C )/(2v)`
D. d) `(v)/(2C)`

Explanation

To find the ratio of the kinetic energy of an electron to that of a photon when both have the same de Broglie wavelength, we can follow these steps:

### Step 1: Write the expression for the kinetic energy of the electron
The kinetic energy (KE) of an electron moving with velocity \( v \) is given by the formula:
\[ KE_e = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the electron.

### Step 2: Write the de Broglie wavelength for the electron
The de Broglie wavelength \( \lambda \) of the electron is given by:
\[ \lambda = \frac{h}{mv} \]
where \( h \) is the Planck constant.

### Step 3: Rearranging the de Broglie wavelength equation
From the de Broglie wavelength equation, we can express \( mv \) in terms of \( \lambda \):
\[ mv = \frac{h}{\lambda} \]

### Step 4: Substitute \( mv \) into the kinetic energy expression
Now, we can substitute \( mv \) into the kinetic energy expression of the electron:
\[ KE_e = \frac{1}{2} mv^2 = \frac{1}{2} \left(\frac{h}{\lambda}\right)v \]

### Step 5: Write the expression for the kinetic energy of the photon
The energy of a photon is given by:
\[ KE_p = h\nu \]
where \( \nu \) is the frequency of the photon. The frequency can be related to the wavelength by:
\[ \nu = \frac{c}{\lambda} \]
where \( c \) is the speed of light.

### Step 6: Substitute \( \nu \) into the photon energy expression
Substituting the expression for \( \nu \) into the photon energy equation gives:
\[ KE_p = h \left(\frac{c}{\lambda}\right) = \frac{hc}{\lambda} \]

### Step 7: Find the ratio of the kinetic energies
Now, we can find the ratio of the kinetic energy of the electron to that of the photon:
\[ \frac{KE_e}{KE_p} = \frac{\frac{1}{2} \left(\frac{h}{\lambda}\right)v}{\frac{hc}{\lambda}} \]

### Step 8: Simplify the ratio
Canceling \( h \) and \( \lambda \) from the numerator and denominator, we get:
\[ \frac{KE_e}{KE_p} = \frac{1}{2} \cdot \frac{v}{c} \]

### Final Answer
Thus, the ratio of the kinetic energy of the electron to that of the photon is:
\[ \frac{KE_e}{KE_p} = \frac{v}{2c} \]