Question

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

• A. $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$
• B. $1$
• C. $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.$
• D. $2$

Answer 1

The correct option is C

$\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.$

Step 1. Given data:

The final kinetic energy of electrons = $2$ X initial kinetic energy of electrons

Step 2. Formula used:

De Broglie wavelength may be given by $\lambda =\frac{h}{p}$……. ..$\left(1\right)$

Where $p$= momentum., $h$= plank's constant.

Step 3. Calculations:

Now, we know momentum $p$= $\sqrt{2m\left(KE\right)}$

Putting the value of $p$ in $\left(1\right)$, we get

$\lambda =\frac{h}{\sqrt{2m\left(KE\right)}}$

Or $\lambda \propto \frac{1}{\sqrt{\left(KE\right)}}$.

So if we double the kinetic energy,

$\lambda \propto \frac{1}{\sqrt{2\left(KE\right)}}$

Or $\lambda \propto \frac{1}{\sqrt{2}}$

Thus, wavelength changes by the factor $\frac{1}{\sqrt{2}}$.

Hence, option (C) is the correct answer.