Question

If the energy of a particle is reduced to half then the percentage increase in the de-Broglie wavelength is about

• A. \(100\%\)
• B. \(61\%\)
• C. \(34\%\)
• D. \(41\%\)

Hint:

If energy becomes half, de Broglie wavelength becomes \(\sqrt{2}\) times. Percentage increase is approximately \(41\%\).

Answer 1

The Correct Option is D

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: \[ \lambda=\frac{h}{\sqrt{2mK}} \] This gives: \[ \lambda\propto \frac{1}{\sqrt{K}} \]

Step 1: Let the initial energy be \(E\).
Initial de Broglie wavelength is: \[ \lambda \] Since: \[ \lambda\propto \frac{1}{\sqrt{E}} \]

Step 2: Energy is reduced to half.
New energy: \[ E'=\frac{E}{2} \] New wavelength: \[ \lambda'\propto \frac{1}{\sqrt{E'}} \] \[ \lambda'\propto \frac{1}{\sqrt{E/2}} \] \[ \lambda'\propto \frac{\sqrt{2}}{\sqrt{E}} \] So: \[ \lambda'=\sqrt{2}\lambda \]

Step 3: Find increase in wavelength.
Increase in wavelength: \[ \Delta\lambda=\lambda'-\lambda \] \[ \Delta\lambda=\sqrt{2}\lambda-\lambda \] \[ \Delta\lambda=(\sqrt{2}-1)\lambda \]

Step 4: Find percentage increase.
Percentage increase: \[ \frac{\Delta\lambda}{\lambda}\times100 \] \[ =(\sqrt{2}-1)\times100 \] Using: \[ \sqrt{2}\approx1.414 \] \[ =(1.414-1)\times100 \] \[ =0.414\times100 \] \[ =41.4\% \] Approximately: \[ 41\% \]

Step 5: Check the options.
Option (A) \(100\%\) is incorrect.
Option (B) \(61\%\) is incorrect.
Option (C) \(34\%\) is incorrect.
Option (D) \(41\%\) is correct. Hence, the correct answer is: \[ \boxed{(D)\ 41\%} \]