Question

The kinetic energy of a charged particle is increased to four times of its initial value. The de Broglie wavelength associated with the particle will :

• A. increase by $100\%$ of its initial value.
• B. increase by $50\%$ of its initial value.
• C. decrease by $25\%$ of its initial value.
• D. decrease by $50\%$ of its initial value.

Hint:

Since $\lambda \propto \frac{1}{\sqrt{K}}$, quadrupling kinetic energy ($K \rightarrow 4K$) halves the wavelength ($\lambda \rightarrow \lambda/2$). Halving means a $50\%$ decrease.

Answer 1

The Correct Option is D

Concept:
• The de Broglie wavelength of a particle of mass $m$ and kinetic energy $K$ is $\lambda = \frac{h}{\sqrt{2mK}}$.

Step 1: Relate new wavelength to initial wavelength
Let initial kinetic energy be $K_1 = K$ and initial wavelength be $\lambda_1 = \frac{h}{\sqrt{2mK}}$.
New kinetic energy $K_2 = 4K$.
New de Broglie wavelength $\lambda_2$ is:
\[ \lambda_2 = \frac{h}{\sqrt{2m(4K)}} = \frac{h}{2\sqrt{2mK}} = \frac{\lambda_1}{2} = 0.5 \lambda_1 \]

Step 2: Calculate percentage change
Percentage change in wavelength is:
\[ \text{Percentage change} = \left( \frac{\lambda_2 - \lambda_1}{\lambda_1} \right) \times 100\% \]
\[ \text{Percentage change} = \left( \frac{0.5\lambda_1 - \lambda_1}{\lambda_1} \right) \times 100\% = -0.5 \times 100\% = -50\% \]
The negative sign indicates a decrease of $50\%$.

Step 3: Conclusion
The de Broglie wavelength decreases by $50\%$ of its initial value, corresponding to option (D).