An electron practically at rest, is initially accelerated through a potential difference of 100 volts. It then has a de Broglie wavlength `=lambda_(1)Å`. It then get retorted through 19 volts and then has a wavelength `lambda_(2)Å` . A further retardation through 32 volts changes the wavelength to `lambda_(3)`. What is the value of `(lambda_(3)-lambda_(2))/(lambda_(1))`?

Options

A. `(20)/(41)`
B. `(10)/(63)`
C. `(20)/(63)`
D. `(10)/(41)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate λ₁
The de Broglie wavelength (λ) of an electron can be calculated using the formula:
\[ \lambda = \frac{12.27 \, \text{Å}}{\sqrt{V}} \]
where \( V \) is the potential difference in volts.

For the first case, where the electron is accelerated through a potential difference of 100 volts:
\[ \lambda_1 = \frac{12.27 \, \text{Å}}{\sqrt{100}} = \frac{12.27 \, \text{Å}}{10} = 1.227 \, \text{Å} \]

### Step 2: Calculate λ₂
The electron is then retorted through 19 volts. The effective potential after retarding is:
\[ V_{\text{effective}} = 100 - 19 = 81 \, \text{volts} \]
Now we can calculate λ₂:
\[ \lambda_2 = \frac{12.27 \, \text{Å}}{\sqrt{81}} = \frac{12.27 \, \text{Å}}{9} = 1.363 \, \text{Å} \]

### Step 3: Calculate λ₃
Next, the electron is further retarded through 32 volts. The effective potential after this retardation is:
\[ V_{\text{effective}} = 81 - 32 = 49 \, \text{volts} \]
Now we can calculate λ₃:
\[ \lambda_3 = \frac{12.27 \, \text{Å}}{\sqrt{49}} = \frac{12.27 \, \text{Å}}{7} = 1.753 \, \text{Å} \]

### Step 4: Calculate \((\lambda_3 - \lambda_2) / \lambda_1\)
Now we can find the value of \((\lambda_3 - \lambda_2) / \lambda_1\):
\[ \lambda_3 - \lambda_2 = 1.753 \, \text{Å} - 1.363 \, \text{Å} = 0.39 \, \text{Å} \]
Now, substituting the values:
\[ \frac{\lambda_3 - \lambda_2}{\lambda_1} = \frac{0.39 \, \text{Å}}{1.227 \, \text{Å}} \approx 0.318 \]

### Final Calculation
To express this in terms of the common factor:
\[ \frac{\lambda_3 - \lambda_2}{\lambda_1} = \frac{12.27 \left(\frac{1}{7} - \frac{1}{9}\right)}{12.27 \left(\frac{1}{10}\right)} = \frac{\frac{1}{7} - \frac{1}{9}}{\frac{1}{10}} = \frac{20}{63} \]

Thus, the final answer is:
\[ \frac{\lambda_3 - \lambda_2}{\lambda_1} = \frac{20}{63} \]