Question

If kinetic energy of free electron is made double, change in de-Broglie wavelength will be........... (A) \(\sqrt{2}\) (B) \((1 / \sqrt{2})\) (C) 2 (D) \((1 / 2)\)

Short answer

The change in de-Broglie wavelength will be \( \frac{1}{\sqrt{2}}\) (B).

Step-by-step solution

Find the initial wavelength

First, let's use the initial kinetic energy to determine the initial de-Broglie wavelength. Using the relationship between kinetic energy and velocity, we can write the equation: $$ \frac{1}{2}mv^2 = K_{1} $$ where \(K_{1}\) is the initial kinetic energy.

Find the final wavelength

Now, let's use the final kinetic energy (which is double the initial kinetic energy) to determine the final de-Broglie wavelength. We can write the equation: $$ \frac{1}{2}mv'^2 = 2K_{1} $$ where \(v'\) is the final velocity.

Find the change in the de-Broglie wavelength

Now, we need to find the ratio \(\frac{\lambda'}{\lambda}\), where \(\lambda'\) is the final de-Broglie wavelength and \(\lambda\) is the initial de-Broglie wavelength. Using the equations from steps 1 and 2, we have: $$ \frac{\lambda'}{\lambda} = \frac{\frac{h}{mv'}}{\frac{h}{mv}} = \frac{v}{v'} $$

Solve the equation for the change in wavelength

We can now solve for the velocity ratio using the kinetic energy equations from steps 1 and 2: $$ v'^2 = 4v^2 $$ $$ \frac{v'}{v} = \frac{1}{\sqrt{2}} $$ Therefore, the change in de-Broglie wavelength will be: $$ \frac{\lambda'}{\lambda} = \boxed{\frac{1}{\sqrt{2}}} \text{ (B)} $$