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. `lambda_(ph) gt lambda_(el)`
B. `lambda_(ph) lt lambda_(el)`
C. `lambda_(ph) = lambda_(el)`
D. `(lambda_(el))/(lambda_(ph)) = C`

Explanation

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

### Step 1: Calculate the wavelength of the photon
The energy of a photon is given by the equation:
\[ E = \frac{hc}{\lambda_{ph}} \]
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 \((3 \times 10^8 \, m/s)\),
- \(\lambda_{ph}\) is the wavelength of the photon.

Rearranging the formula to find \(\lambda_{ph}\):
\[ \lambda_{ph} = \frac{hc}{E} \]

Substituting the values:
\[ \lambda_{ph} = \frac{(6.626 \times 10^{-34} \, J \cdot s)(3 \times 10^8 \, m/s)}{10^{-20} \, J} \]
Calculating this gives:
\[ \lambda_{ph} = \frac{1.9878 \times 10^{-25}}{10^{-20}} = 1.9878 \times 10^{-5} \, m = 1.9878 \times 10^{-5} \, m = 1.988 \times 10^{-5} \, m \]

### Step 2: Calculate the wavelength of the electron
For an electron, we use the de Broglie wavelength formula:
\[ \lambda_{el} = \frac{h}{p} \]
Where \(p\) is the momentum of the electron. The momentum can be expressed in terms of energy:
\[ p = \sqrt{2mE} \]
Thus, we can rewrite the wavelength as:
\[ \lambda_{el} = \frac{h}{\sqrt{2mE}} \]

Substituting the values:
- Mass of the electron \(m \approx 9.11 \times 10^{-31} \, kg\),
- Energy \(E = 10^{-20} \, J\).

Calculating the momentum:
\[ p = \sqrt{2(9.11 \times 10^{-31} \, kg)(10^{-20} \, J)} = \sqrt{1.822 \times 10^{-50}} \approx 1.351 \times 10^{-25} \, kg \cdot m/s \]

Now substituting this back into the de Broglie wavelength formula:
\[ \lambda_{el} = \frac{6.626 \times 10^{-34} \, J \cdot s}{1.351 \times 10^{-25} \, kg \cdot m/s} \approx 4.90 \times 10^{-9} \, m \]

### Step 3: Compare the wavelengths
Now we have:
- \(\lambda_{ph} \approx 1.988 \times 10^{-5} \, m\)
- \(\lambda_{el} \approx 4.90 \times 10^{-9} \, m\)

Since \(1.988 \times 10^{-5} \, m > 4.90 \times 10^{-9} \, m\), we can conclude that:
\[ \lambda_{ph} > \lambda_{el} \]

### Conclusion
The correct statement is that the wavelength associated with the photon is greater than that associated with the electron.