Question

The ratio of the respective de Broglie wavelengths of two particles with kinetic energy of 0.02 eV and 2 eV is ________.

• A. 1:1
• B. 10:1
• C. 1:10
• D. $1:\sqrt{10}$
• E. $\sqrt{10}:1$

Hint:

Higher kinetic energy means a shorter de Broglie wavelength.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
This question deals with the de Broglie wavelength, which associates a wavelength with any moving particle. The wavelength depends on the particle's momentum. We need to find the relationship between de Broglie wavelength and kinetic energy and then calculate the ratio for the two given energies.
Step 2: Key Formula or Approach
The de Broglie wavelength (\(\lambda\)) is given by:
\[ \lambda = \frac{h}{p} \] where \(h\) is Planck's constant and \(p\) is the momentum of the particle.
The kinetic energy (\(K\)) of a non-relativistic particle is related to its momentum by:
\[ K = \frac{p^2}{2m} \implies p = \sqrt{2mK} \] Substituting this into the de Broglie wavelength formula gives:
\[ \lambda = \frac{h}{\sqrt{2mK}} \] This shows that for a given particle (constant mass \(m\)), the wavelength is inversely proportional to the square root of its kinetic energy:
\[ \lambda \propto \frac{1}{\sqrt{K}} \] Step 3: Detailed Explanation
1. Set up the ratio.
Let the two particles have kinetic energies \(K_1\) and \(K_2\), and corresponding wavelengths \(\lambda_1\) and \(\lambda_2\). We assume the particles are identical (same mass \(m\)).
The ratio of their wavelengths is:
\[ \frac{\lambda_1}{\lambda_2} = \frac{h/\sqrt{2mK_1}}{h/\sqrt{2mK_2}} = \frac{\sqrt{2mK_2}}{\sqrt{2mK_1}} = \sqrt{\frac{K_2}{K_1}} \] 2. List the given kinetic energies.
- \(K_1 = 0.02 \text{ eV}\)
- \(K_2 = 2 \text{ eV}\)
3. Calculate the ratio of wavelengths.
\[ \frac{\lambda_1}{\lambda_2} = \sqrt{\frac{2}{0.02}} \] \[ \frac{\lambda_1}{\lambda_2} = \sqrt{\frac{200}{2}} = \sqrt{100} = 10 \] So, the ratio \(\lambda_1 : \lambda_2\) is 10 : 1.
Step 4: Final Answer
The ratio of the de Broglie wavelengths is 10 : 1.