The de Broglie wavelength of an electron moving with a velocity of `1.5xx10^(8)ms^(-1)` is equal to that of a photon find the ratio of the kinetic energy of the photon to that of the electron.

Options

A. 2
B. 4
C. `(1)/(2)`
D. `(1)/(4)`

Answer

To solve the problem of finding the ratio of the kinetic energy of a photon to that of an electron, we will follow these steps:

### Step 1: Understand the de Broglie wavelength
The de Broglie wavelength (\( \lambda \)) of a particle is given by the formula:
\[ \lambda = \frac{h}{mv} \]
where:
- \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, \text{Js} \)),
- \( m \) is the mass of the particle,
- \( v \) is the velocity of the particle.

### Step 2: Kinetic energy of the photon
The kinetic energy (\( KE \)) of a photon can be expressed as:
\[ KE_{\text{photon}} = h \nu \]
where \( \nu \) (nu) is the frequency of the photon. The frequency can be related to the wavelength by the equation:
\[ \nu = \frac{c}{\lambda} \]
where \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)). Therefore, we can rewrite the kinetic energy of the photon as:
\[ KE_{\text{photon}} = h \frac{c}{\lambda} \]

### Step 3: Kinetic energy of the electron
The kinetic energy of an electron is given by:
\[ KE_{\text{electron}} = \frac{1}{2} mv^2 \]

### Step 4: Setting the wavelengths equal
According to the problem, the de Broglie wavelength of the electron is equal to the wavelength of the photon:
\[ \lambda_{\text{electron}} = \lambda_{\text{photon}} \]
Thus, we can set the two expressions for wavelength equal to each other:
\[ \frac{h}{mv} = \lambda \]

### Step 5: Substitute the wavelength into the kinetic energy ratio
Now we can find the ratio of the kinetic energies:
\[ \frac{KE_{\text{photon}}}{KE_{\text{electron}}} = \frac{h \frac{c}{\lambda}}{\frac{1}{2} mv^2} \]
Substituting \( \lambda = \frac{h}{mv} \):
\[ \frac{KE_{\text{photon}}}{KE_{\text{electron}}} = \frac{h \frac{c}{\frac{h}{mv}}}{\frac{1}{2} mv^2} \]
This simplifies to:
\[ \frac{KE_{\text{photon}}}{KE_{\text{electron}}} = \frac{h \cdot c \cdot mv}{h \cdot \frac{1}{2} mv^2} \]
Cancelling \( h \) and \( mv \):
\[ \frac{KE_{\text{photon}}}{KE_{\text{electron}}} = \frac{2c}{v} \]

### Step 6: Substitute the known values
Given that \( v = 1.5 \times 10^8 \, \text{m/s} \) and \( c = 3 \times 10^8 \, \text{m/s} \):
\[ \frac{KE_{\text{photon}}}{KE_{\text{electron}}} = \frac{2 \cdot 3 \times 10^8}{1.5 \times 10^8} \]
This simplifies to:
\[ \frac{KE_{\text{photon}}}{KE_{\text{electron}}} = \frac{6}{1.5} = 4 \]

### Conclusion
Thus, the ratio of the kinetic energy of the photon to that of the electron is:
\[ \frac{KE_{\text{photon}}}{KE_{\text{electron}}} = 4 \]

### Final Answer
The answer is option B: 4.

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