Question

If the kinetic energy of a particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is

• A. \(75%\)
• B. \(25%\)
• C. \(50%\)
• D. \(5%\)

Hint:

De-Broglie wavelength varies inversely with the square root of kinetic energy.

Answer 1

The Correct Option is A

Step 1: Relation between de-Broglie wavelength and kinetic energy.
\[ \lambda = \frac{h}{\sqrt{2mK}} \Rightarrow \lambda \propto \frac{1}{\sqrt{K}} \]

Step 2: Apply increase in kinetic energy.
If \(K' = 16K\), then \[ \lambda' = \frac{\lambda}{\sqrt{16}} = \frac{\lambda}{4} \]

Step 3: Calculate percentage change.
\[ %\text{ decrease} = \frac{\lambda - \lambda'}{\lambda}\times100 = \frac{3}{4}\times100 = 75% \]