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 need to calculate the de Broglie wavelengths of the electron at different stages of its acceleration and retardation. We will use the relationship between the de Broglie wavelength and the kinetic energy of the electron, which is influenced by the potential difference it has been accelerated or retarded through.

### Step-by-Step Solution:

1. **Calculate the initial kinetic energy after acceleration through 100 V:**
\[ KE_1 = e \cdot V_1 = (1.6 \times 10^{-19} \, \text{C}) \cdot (100 \, \text{V}) = 1.6 \times 10^{-17} \, \text{J} \]

2. **Relate kinetic energy to momentum:**
The kinetic energy can also be expressed in terms of momentum \( p \):
\[ KE = \frac{p^2}{2m} \implies p = \sqrt{2m \cdot KE} \]
where \( m \) is the mass of the electron, \( m = 9.1 \times 10^{-31} \, \text{kg} \).

3. **Calculate momentum \( p_1 \):**
\[ p_1 = \sqrt{2 \cdot (9.1 \times 10^{-31} \, \text{kg}) \cdot (1.6 \times 10^{-17} \, \text{J})} \]

4. **Calculate the de Broglie wavelength \( \lambda_1 \):**
The de Broglie wavelength is given by:
\[ \lambda = \frac{h}{p} \]
where \( h = 6.626 \times 10^{-34} \, \text{Js} \). Thus,
\[ \lambda_1 = \frac{h}{p_1} \]

5. **Calculate the new kinetic energy after retardation of 19 V:**
\[ KE_2 = e \cdot V_2 = e \cdot (100 - 19) = (1.6 \times 10^{-19} \, \text{C}) \cdot (81 \, \text{V}) = 1.296 \times 10^{-17} \, \text{J} \]

6. **Calculate momentum \( p_2 \):**
\[ p_2 = \sqrt{2m \cdot KE_2} \]

7. **Calculate the de Broglie wavelength \( \lambda_2 \):**
\[ \lambda_2 = \frac{h}{p_2} \]

8. **Calculate the new kinetic energy after further retardation of 32 V:**
\[ KE_3 = e \cdot V_3 = e \cdot (81 - 32) = (1.6 \times 10^{-19} \, \text{C}) \cdot (49 \, \text{V}) = 7.84 \times 10^{-18} \, \text{J} \]

9. **Calculate momentum \( p_3 \):**
\[ p_3 = \sqrt{2m \cdot KE_3} \]

10. **Calculate the de Broglie wavelength \( \lambda_3 \):**
\[ \lambda_3 = \frac{h}{p_3} \]

11. **Calculate the final expression:**
\[ \frac{\lambda_3 - \lambda_2}{\lambda_1} \]

### Final Calculation:
Now, substituting the values calculated in the above steps will give us the final answer.