Question

The kinetic energy of a free electron increases to 3 times the previous K.E. The ratio of new de-Broglie wavelength to previous de-Broglie wavelength is

• A. \(\frac{1}{\sqrt{3}}\)
• B. \(\frac{1}{3}\)
• C. \(3\)
• D. \(\sqrt{3}\)

Hint:

\(\lambda=\dfrac{h}{\sqrt{2mK}}\), so \(\lambda\propto\dfrac{1}{\sqrt K}\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The de Broglie wavelength of a particle of kinetic energy \(K\) is \(\lambda=\dfrac h p=\dfrac{h}{\sqrt{2mK}}\).

Step 2: Detailed Explanation
If \(K\) becomes \(3K\), then
\[ \frac{\lambda'}{\lambda}=\sqrt{\frac{K}{3K}}=\frac1{\sqrt3} \]

Final Answer:
The wavelength ratio is \(\frac1{\sqrt3}\), option (A). \[ \boxed{\dfrac1{\sqrt3}\ \text{(A)}} \]