Question

An electron travelling with velocity \(v\) in free space enters a medium where its velocity is reduced by \(20\%\). The de-Broglie wavelength of the electron in the medium is \(\alpha \lambda_0\), where \(\lambda_0\) is the wavelength in free space. Find the value of \(\alpha\).

• A. \(1.25\)
• B. \(1.5\)
• C. \(0.8\)
• D. \(0.125\)

Hint:

For de-Broglie wavelength: \[ \lambda = \frac{h}{mv} \] If velocity decreases, wavelength increases proportionally.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The de-Broglie wavelength of a moving particle is inversely proportional to its momentum.
When the velocity of the electron decreases, its wavelength increases proportionally.
Step 2: Key Formula or Approach:
The de-Broglie wavelength is given by $\lambda = \frac{h}{mv}$.
This implies the relation $\lambda \propto \frac{1}{v}$.
Step 3: Detailed Explanation:
Let the initial velocity in free space be $v_0$.
The initial wavelength is $\lambda_0 \propto \frac{1}{v_0}$.
When it enters the medium, its velocity is reduced by 20%.
The new velocity $v$ is:
\[ v = v_0 - 0.20 v_0 = 0.8 v_0 \]
The new de-Broglie wavelength $\lambda$ in the medium is:
\[ \lambda \propto \frac{1}{0.8v_0} \]
Taking the ratio of the new wavelength to the original wavelength:
\[ \frac{\lambda}{\lambda_0} = \frac{v_0}{v} = \frac{v_0}{0.8v_0} = \frac{1}{0.8} = 1.25 \]
Thus, $\lambda = 1.25 \lambda_0$.
Comparing this with $\lambda = \alpha\lambda_0$, we get $\alpha = 1.25$.
Step 4: Final Answer:
The value of $\alpha$ is 1.25.