Question

Two electrons are moving with different kinetic energies. If the first electron is moving with kinetic energy \(k\) and the second electron with kinetic energy \(4k\), the ratio of their de Broglie wavelengths respectively will be:

• A. \(1:2\)
• B. \(1:4\)
• C. \(2:1\)
• D. \(4:1\)

Hint:

For the same particle, de Broglie wavelength varies inversely as the square root of kinetic energy: \[ \lambda \propto \frac{1}{\sqrt{K}} \]

Answer 1

The Correct Option is C

Concept:
The de Broglie wavelength of a particle is: \[ \lambda=\frac{h}{p} \] For a non-relativistic particle, \[ K=\frac{p^2}{2m} \] So, \[ p=\sqrt{2mK} \] Therefore, \[ \lambda=\frac{h}{\sqrt{2mK}} \] Thus, \[ \lambda \propto \frac{1}{\sqrt{K}} \]

Step 1: Write the kinetic energies.
For first electron: \[ K_1=k \] For second electron: \[ K_2=4k \]

Step 2: Use relation between wavelength and kinetic energy.
\[ \frac{\lambda_1}{\lambda_2}=\sqrt{\frac{K_2}{K_1}} \] Substitute the values: \[ \frac{\lambda_1}{\lambda_2}=\sqrt{\frac{4k}{k}} \] \[ =\sqrt{4} \] \[ =2 \]

Step 3: Write the ratio.
\[ \lambda_1:\lambda_2=2:1 \]

Step 4: Final conclusion.
Hence, the ratio of their de Broglie wavelengths is: \[ \boxed{2:1} \]