Question

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

• A. \(40 \%\)
• B. \(60 \%\)
• C. \(75 \%\)
• D. \(80 \%\)

Hint:

Wavelength is inversely proportional to the square root of kinetic energy.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
\(\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}\), so \(\lambda \propto \frac{1}{\sqrt K}\).

Step 2: Key Formula or Approach:
New kinetic energy \(K' = 16K\).

Step 3: Detailed Explanation:
\(\lambda' = \frac{\lambda}{\sqrt{16}} = \frac\lambda4\).
Change \(= \lambda - \frac\lambda4 = \frac{3\lambda}{4}\), which is a decrease of \(75\%\).
\[ \text{Percentage change} = 75\%\ \text{(decrease)} \]

Final Answer:
The wavelength decreases by \(75\%\), option (C). \[ \boxed{75\%} \]