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Step-by-Step Solution
Step 1: Identify the Relevant Concept
The de Broglie wavelength $ \lambda $ of an electron accelerated by a potential difference $ V $ is inversely proportional to the square root of the voltage. Mathematically,
$$
\lambda \propto \frac{1}{\sqrt{V}}.
$$
A commonly used approximate formula for the de Broglie wavelength (in angstroms) is:
$$
\lambda = \frac{1.227}{\sqrt{V}},
$$
where $ V $ is in volts.
Step 2: Express the Relationship Between Wavelength and Voltage
From $ \lambda \propto \frac{1}{\sqrt{V}} $, if the voltage changes from an initial value $ V_1 $ to a final value $ V_2 $, we have
$$
\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{V_2}{V_1}}.
$$
Step 3: Substitute the Given Values
Given:
$$
V_1 = 25\,\text{kV} = 25 \times 10^3 \,\text{V}, \quad V_2 = 100\,\text{kV} = 100 \times 10^3 \,\text{V}.
$$
Hence,
$$
\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{100 \times 10^3}{25 \times 10^3}}
= \sqrt{\frac{100}{25}}
= \sqrt{4}
= 2.
$$
Therefore,
$$
\lambda_2 = \frac{\lambda_1}{2}.
$$
Step 4: Conclusion
Since $ \lambda_2 = \frac{\lambda_1}{2} $, the de Broglie wavelength decreases by a factor of 2 when the accelerating voltage is increased from 25 kV to 100 kV. Hence, the correct answer is that the wavelength decreases by 2 times.
