Photon and electron are given same energy `(10^(-20) J)`. Wavelength associated with photon and electron are `lambda_(ph)` and `lambda_(el)` then correct statement will be

Options

A. `lamda_(ph) = lamda_(e)`
B. `lamda_(ph) lt lamda_(e)`
C. `lamda_(e)//lamda_(ph) = c`
D. `lamda_(ph) gt lamda_(e)`

Explanation

To solve the problem, we need to find the wavelengths associated with a photon and an electron when both have the same energy of \( 10^{-20} \, J \).

### Step 1: Understand the Energy of the Photon
The energy of a photon is given by the equation:
\[ E = \frac{hc}{\lambda} \]
where:
- \( E \) is the energy of the photon,
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, J \cdot s \)),
- \( c \) is the speed of light (\( 3 \times 10^8 \, m/s \)),
- \( \lambda \) is the wavelength of the photon.

Rearranging this equation to find the wavelength of the photon (\( \lambda_{ph} \)):
\[ \lambda_{ph} = \frac{hc}{E} \]

### Step 2: Calculate the Wavelength of the Photon
Substituting the values into the equation:
\[ \lambda_{ph} = \frac{(6.626 \times 10^{-34} \, J \cdot s)(3 \times 10^8 \, m/s)}{10^{-20} \, J} \]
Calculating this gives:
\[ \lambda_{ph} = \frac{1.9878 \times 10^{-25}}{10^{-20}} = 1.9878 \times 10^{-5} \, m = 1.9878 \times 10^{-5} \, m = 1.9878 \times 10^{-5} \, m \]

### Step 3: Understand the Energy of the Electron
For an electron, the energy can be related to its momentum \( p \) using the equation:
\[ E = \frac{p^2}{2m} \]
Rearranging to find the momentum:
\[ p = \sqrt{2mE} \]
where \( m \) is the mass of the electron (\( 9.11 \times 10^{-31} \, kg \)).

### Step 4: Calculate the Momentum of the Electron
Substituting the values:
\[ p = \sqrt{2 \times (9.11 \times 10^{-31} \, kg) \times (10^{-20} \, J)} \]
Calculating this gives:
\[ p = \sqrt{1.822 \times 10^{-50}} = 1.351 \times 10^{-25} \, kg \cdot m/s \]

### Step 5: Calculate the Wavelength of the Electron
Using de Broglie's wavelength formula:
\[ \lambda_{el} = \frac{h}{p} \]
Substituting the values:
\[ \lambda_{el} = \frac{6.626 \times 10^{-34} \, J \cdot s}{1.351 \times 10^{-25} \, kg \cdot m/s} \]
Calculating this gives:
\[ \lambda_{el} = 4.905 \times 10^{-9} \, m \]

### Step 6: Compare the Wavelengths
Now we have:
- \( \lambda_{ph} \approx 1.9878 \times 10^{-5} \, m \)
- \( \lambda_{el} \approx 4.905 \times 10^{-9} \, m \)

Since \( \lambda_{ph} > \lambda_{el} \), we conclude that:
\[ \lambda_{ph} > \lambda_{el} \]

### Conclusion
The correct statement is that the wavelength associated with the photon (\( \lambda_{ph} \)) is greater than the wavelength associated with the electron (\( \lambda_{el} \)).