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:

Remember the important relation \[ \lambda\propto\frac{1}{\sqrt{K}}. \] If kinetic energy becomes \(n\) times, the wavelength becomes \(1/\sqrt{n}\) times.

Answer 1

The Correct Option is D

Concept: According to de Broglie's hypothesis, \[ \lambda=\frac{h}{p}, \] where \(h\) is Planck's constant and \(p\) is the momentum of the particle. For a non-relativistic particle, \[ K=\frac{p^2}{2m}. \] Therefore, \[ p=\sqrt{2mK}. \] Substituting into the de Broglie relation, \[ \lambda=\frac{h}{\sqrt{2mK}}. \] Hence, \[ \lambda\propto\frac{1}{\sqrt{K}}. \]

Step 1: Write the relation between wavelength and kinetic energy. Since \[ \lambda\propto\frac{1}{\sqrt{K}}, \] we can write \[ \frac{\lambda_2}{\lambda_1} = \sqrt{\frac{K_1}{K_2}}. \]

Step 2: Use the given information. The final kinetic energy is four times the initial kinetic energy. Therefore, \[ K_2=4K_1. \] Substituting, \[ \frac{\lambda_2}{\lambda_1} = \sqrt{\frac{K_1}{4K_1}}. \] \[ \frac{\lambda_2}{\lambda_1} = \sqrt{\frac{1}{4}}. \] \[ \frac{\lambda_2}{\lambda_1} = \frac{1}{2}. \] Thus, \[ \lambda_2=\frac{\lambda_1}{2}. \]

Step 3: Determine the percentage decrease. Initial wavelength \[ =\lambda_1. \] Final wavelength \[ =\frac{\lambda_1}{2}. \] Decrease \[ =\lambda_1-\frac{\lambda_1}{2} =\frac{\lambda_1}{2}. \] Percentage decrease \[ = \frac{\frac{\lambda_1}{2}}{\lambda_1}\times100 = 50\%. \]

Step 4: Write the final answer. The de Broglie wavelength becomes half of its original value. Hence, it decreases by \[ 50\%. \] Therefore, \[ \boxed{\text{(D)}} \] is the correct answer.