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 every moving particle has an associated wave nature. The wavelength (\( \lambda \)) is inversely proportional to the square root of the particle's kinetic energy ($K$).
Step 2: Key Formula or Approach:
\[ \lambda = \frac{h}{\sqrt{2mK}} \implies \lambda \propto \frac{1}{\sqrt{K}} \] Therefore: \[ \frac{\lambda_2}{\lambda_1} = \sqrt{\frac{K_1}{K_2}} \]
Step 3: Detailed Explanation:
1. Given: \( \lambda_1 = 2000 \) Å and \( K_1 = 1 \) eV.
2. New kinetic energy \( K_2 = 1 \) MeV = \( 10^6 \) eV.
3. Substitute the values into the ratio formula: \[ \frac{\lambda_2}{2000} = \sqrt{\frac{1}{10^6}} \] \[ \frac{\lambda_2}{2000} = \frac{1}{1000} \] \[ \lambda_2 = \frac{2000}{1000} = 2 \text{ Å} \]
Step 4: Final Answer:
The new de Broglie wavelength is 2 Å.