Question

The additional energy that should be given to an electron to reduce its de-Broglie wavelength from 1 nm to 0.5 nm is \(x\) times the initial kinetic energy. The value of \(x\) is

• A. \(0.5\)
• B. \(2\)
• C. \(3\)
• D. \(4\)

Hint:

Kinetic energy varies as the inverse square of the de Broglie wavelength, so halving the wavelength multiplies the energy by four.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The de Broglie wavelength of an electron is \(\lambda=\dfrac{h}{\sqrt{2mK}}\), so the kinetic energy is \(K=\dfrac{h^{2}}{2m\lambda^{2}}\), that is \(K\propto\dfrac{1}{\lambda^{2}}\).

Step 2: Calculation
The wavelength falls from 1 nm to 0.5 nm, so \(\lambda\) becomes half. Then \(K'=\left(\dfrac{1}{1/2}\right)^{2}K=4K\).

Step 3: Extra energy
Additional energy \(=K'-K=4K-K=3K\).

Final Answer:
The additional energy is 3 times the initial kinetic energy, so \(x=3\). \[ \boxed{\text{(C) }3} \]