Question

The de Broglie wavelength and kinetic energy of a particle is 2000 Å and 1 eV respectively. If its kinetic energy becomes 1 MeV, then its de Broglie wavelength is

• A. 2 Å
• B. 1 Å
• C. 4 Å
• D. 10 Å
• E. 5 Å

Hint:

If the kinetic energy increases by a factor of \( X \), the wavelength decreases by a factor of \( \sqrt{X} \). Here, energy increased \( 10^6 \) times, so wavelength decreased \( \sqrt{10^6} = 1000 \) times.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The de Broglie hypothesis states that a moving particle has an associated wave nature. The wavelength of this matter wave (\( \lambda \)) is inversely proportional to the particle's momentum (\( p \)).
Step 2: Key Formula or Approach:
The de Broglie wavelength is given by: \[ \lambda = \frac{h}{p} \] The kinetic energy (\( K \)) and momentum (\( p \)) of a non-relativistic particle are related by \( K = \frac{p^2}{2m} \), which gives \( p = \sqrt{2mK} \). Substituting this into the wavelength formula yields: \[ \lambda = \frac{h}{\sqrt{2mK}} \] For the same particle, mass \( m \) is constant, so \( \lambda \propto \frac{1}{\sqrt{K}} \).
Step 3: Detailed Explanation:
Let the initial state be 1 and the final state be 2.
Given values:
\( \lambda_1 = 2000 \text{ \AA} \)
\( K_1 = 1 \text{ eV} \)
\( K_2 = 1 \text{ MeV} = 10^6 \text{ eV} \)
Using the proportional relationship, we set up the ratio: \[ \frac{\lambda_2}{\lambda_1} = \sqrt{\frac{K_1}{K_2}} \] Substitute the known values into the equation: \[ \frac{\lambda_2}{2000 \text{ \AA}} = \sqrt{\frac{1 \text{ eV}}{10^6 \text{ eV}}} \] \[ \frac{\lambda_2}{2000 \text{ \AA}} = \sqrt{10^{-6}} \] \[ \frac{\lambda_2}{2000 \text{ \AA}} = 10^{-3} \] Now, solve for the final wavelength \( \lambda_2 \): \[ \lambda_2 = 2000 \text{ \AA} \times 10^{-3} \] \[ \lambda_2 = 2 \text{ \AA} \]
Step 4: Final Answer:
The new de Broglie wavelength is \( 2 \text{ \AA} \).