There are two very small natural particle which are held fixed at a separation of 1 m from each other . Using some mechanism , one electron is transferred from particle to the other and electric potential energy lost in this process is converted into a photon of wavelength `lamda_1 `. When one more electron is transferred in the same direction , the wavelength of photon emitted is ` lamda_2` . If `lamda_1 -lamda_2=5^pxx23^q` calculate the value of pq.
`{:(0,1,2,3,4,5,6,7,8,9):}`

Explanation

To solve the problem, we need to calculate the difference in wavelengths of the photons emitted when two electrons are transferred between two particles. We'll follow these steps:

### Step 1: Understand the Electric Potential Energy Change
When an electron is transferred from one particle to another, the electric potential energy (U) changes. The formula for electric potential energy between two charges is given by:

\[ U = -\frac{k \cdot q_1 \cdot q_2}{d} \]

where \( k = \frac{1}{4 \pi \epsilon_0} \), \( q_1 \) and \( q_2 \) are the charges, and \( d \) is the distance between them.

### Step 2: Calculate U2 for the First Electron Transfer
Initially, both particles have zero charge. After transferring one electron (charge \( e = 1.6 \times 10^{-19} \) C), the charges become \( +e \) and \( -e \). The potential energy after the first transfer (U2) is:

\[ U_2 = -\frac{k \cdot e \cdot (-e)}{1} = \frac{k \cdot e^2}{1} \]

Substituting \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \):

\[ U_2 = 9 \times 10^9 \cdot (1.6 \times 10^{-19})^2 = 23.04 \times 10^{-29} \, \text{J} \]

### Step 3: Calculate U3 for the Second Electron Transfer
When a second electron is transferred, the charges become \( +2e \) and \( -2e \). The potential energy after the second transfer (U3) is:

\[ U_3 = -\frac{k \cdot (2e) \cdot (-2e)}{1} = \frac{4k \cdot e^2}{1} \]

Substituting the value:

\[ U_3 = 4 \cdot 23.04 \times 10^{-29} = 92.16 \times 10^{-29} \, \text{J} \]

### Step 4: Calculate the Wavelengths
The energy of the emitted photon is given by:

\[ E = \frac{hc}{\lambda} \]

where \( h = 6.63 \times 10^{-34} \, \text{J s} \) and \( c = 3 \times 10^8 \, \text{m/s} \).

#### For λ1:
The energy difference when the first electron is transferred is:

\[ U_1 - U_2 = 0 - (-23.04 \times 10^{-29}) = 23.04 \times 10^{-29} \, \text{J} \]

Setting this equal to the photon energy:

\[ 23.04 \times 10^{-29} = \frac{6.63 \times 10^{-34} \cdot 3 \times 10^8}{\lambda_1} \]

Solving for \( \lambda_1 \):

\[ \lambda_1 = \frac{6.63 \times 10^{-34} \cdot 3 \times 10^8}{23.04 \times 10^{-29}} \approx 863 \, \text{m} \]

#### For λ2:
The energy difference when the second electron is transferred is:

\[ U_2 - U_3 = -23.04 \times 10^{-29} - (-92.16 \times 10^{-29}) = 69.12 \times 10^{-29} \, \text{J} \]

Setting this equal to the photon energy:

\[ 69.12 \times 10^{-29} = \frac{6.63 \times 10^{-34} \cdot 3 \times 10^8}{\lambda_2} \]

Solving for \( \lambda_2 \):

\[ \lambda_2 = \frac{6.63 \times 10^{-34} \cdot 3 \times 10^8}{69.12 \times 10^{-29}} \approx 288 \, \text{m} \]

### Step 5: Calculate the Difference in Wavelengths
Now, we find the difference:

\[ \lambda_1 - \lambda_2 = 863 - 288 = 575 \, \text{m} \]

### Step 6: Express the Difference in the Given Form
We need to express 575 in the form \( 5^p \cdot 23^q \):

\[ 575 = 5^2 \cdot 23^1 \]

### Step 7: Identify p and q
From the expression \( 5^p \cdot 23^q \):

- \( p = 2 \)
- \( q = 1 \)

### Final Calculation
The value of \( pq \) is:

\[ pq = 2 \cdot 1 = 2 \]

### Conclusion
The final answer is:

\[ \boxed{2} \]