Question

If the kinetic energy of a free electron doubles, its de Broglie wavelength changes by the factor

• A. \(2\)
• B. \(1/2\)
• C. \(\sqrt{2}\)
• D. \(1/\sqrt{2}\)

Hint:

For a moving particle: \[ \lambda\propto \frac{1}{\sqrt{K}} \] So if kinetic energy doubles, wavelength becomes \(\frac{1}{\sqrt{2}}\) times.

Answer 1

The Correct Option is D

Concept:
The de Broglie wavelength of a moving particle is: \[ \lambda=\frac{h}{p} \] where, \[ \lambda=\text{de Broglie wavelength} \] \[ h=\text{Planck's constant} \] \[ p=\text{momentum} \] For a non-relativistic particle: \[ K=\frac{p^2}{2m} \] So: \[ p=\sqrt{2mK} \] Therefore: \[ \lambda=\frac{h}{\sqrt{2mK}} \] This means: \[ \lambda\propto \frac{1}{\sqrt{K}} \]

Step 1: Write the relation between wavelength and kinetic energy.
Since: \[ \lambda\propto \frac{1}{\sqrt{K}} \] If kinetic energy increases, de Broglie wavelength decreases.

Step 2: Let the initial kinetic energy be \(K\).
Initial wavelength: \[ \lambda=\frac{h}{\sqrt{2mK}} \] Now kinetic energy doubles: \[ K'=2K \]

Step 3: Find the new wavelength.
New wavelength: \[ \lambda'=\frac{h}{\sqrt{2mK'}} \] Substitute: \[ K'=2K \] \[ \lambda'=\frac{h}{\sqrt{2m(2K)}} \] \[ \lambda'=\frac{h}{\sqrt{2}\sqrt{2mK}} \] \[ \lambda'=\frac{\lambda}{\sqrt{2}} \]

Step 4: Find the factor.
So the wavelength changes by the factor: \[ \frac{1}{\sqrt{2}} \]

Step 5: Check the options.
Option (A) \(2\) is incorrect because wavelength decreases, not doubles.
Option (B) \(1/2\) is incorrect.
Option (C) \(\sqrt{2}\) is incorrect because wavelength does not increase.
Option (D) \(1/\sqrt{2}\) is correct. Hence, the correct answer is: \[ \boxed{(D)\ \frac{1}{\sqrt{2}}} \]