Question

The de Broglie wavelength for an electron accelerated through the potential difference of \(V_1\) volt is \( \lambda_1 \). When the potential difference is changed to \(V_2\) volt, the associated de Broglie wavelength is increased by \(50%\). If \( \left(\frac{V_1}{V_2}\right)=\frac{9}{\alpha} \), then the value of \( \alpha \) is _____.

Answer 1

Correct Answer: 4

Step 1: Understanding the Question:
We need to find the ratio of potentials based on the change in the de Broglie wavelength of an electron.
Step 2: Key Formula or Approach:
The de Broglie wavelength \(\lambda\) of an electron accelerated through potential \(V\) is:
\[ \lambda = \frac{h}{\sqrt{2mqV}} \Rightarrow \lambda \propto \frac{1}{\sqrt{V}} \Rightarrow V \propto \frac{1}{\lambda^2} \]
Step 3: Detailed Explanation:
1. Initial state: wavelength = \(\lambda_1\), potential = \(V_1\).
2. Final state: wavelength = \(\lambda_2 = \lambda_1 + 0.5\lambda_1 = 1.5\lambda_1\), potential = \(V_2\).
3. Using the inverse square relation:
\[ \frac{V_1}{V_2} = \left( \frac{\lambda_2}{\lambda_1} \right)^2 = (1.5)^2 \]
\[ \frac{V_1}{V_2} = 2.25 = \frac{225}{100} = \frac{9}{4} \]
4. Comparing with \(\frac{V_1}{V_2} = \frac{9}{\alpha}\):
\[ \frac{9}{4} = \frac{9}{\alpha} \Rightarrow \alpha = 4 \]
Step 4: Final Answer:
The value of \(\alpha\) is 4.