Question

If the kinetic energy of a particle is doubled, de Broglie wavelength becomes:

• A. $2$ times
• B. $4$ times
• C. $\sqrt{2}$ times
• D. $\frac{1}{\sqrt{2}}$ times

Answer 1

The correct option is D

$\frac{1}{\sqrt{2}}$ times

Step 1: Given data

Kinetic energy of a particle is doubled

$KE\text{'}=2KE$

Step 2: Formula used

We know, kinetic energy is given by,

$KE=\frac{1}{2}m{v}^2$

momentum,

$p=mv$

so momentum in terms of kinetic energy,

$$\begin{gathered} KE=\frac{1}{2}m{v}^2 \\ \Rightarrow mv=\sqrt{2mKE} \\ \Rightarrow p=\sqrt{2mKE} \end{gathered}$$

De - Broglie wavelength,

$\lambda =\frac{h}{p}$

where $h$ is planks constant

Step 3: Find the relation between kinetic energy and de-Broglie wavelength

Momentum in terms of kinetic energy,

$p=\sqrt{2mKE}$

de-Broglie wavelength,

$$\begin{gathered} \lambda =\frac{h}{p} \\ \Rightarrow \lambda \propto \frac{1}{p} \\ \Rightarrow \lambda \propto \frac{1}{\sqrt{KE}}\left(p=\sqrt{2mKE}\right) \end{gathered}$$

since $m$ is constant

Step 4: Finding the change in de-Broglie wavelength when the kinetic energy is doubled

$$\begin{gathered} \Rightarrow \lambda \propto \frac{1}{\sqrt{KE}} \\ \Rightarrow \lambda \text{'}\propto \frac{1}{\sqrt{KE\text{'}}} \\ \Rightarrow \lambda \text{'}\propto \frac{1}{\sqrt{2KE}}\left(\because KE\text{'}=2KE\right) \\ \Rightarrow \lambda \text{'}\propto \frac{1}{\sqrt{2}}\lambda \end{gathered}$$

Since wavelength is inversely proportional to the square root of kinetic energy, when the kinetic energy is doubled, the de-Broglie wavelength becomes$\frac{1}{\sqrt{2}}$ times the original wavelength.

Option D is correct.