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: Use momentum
$K=\dfrac{p^2}{2m}$, so tripling $K$ multiplies $p$ by $\sqrt3$.

Step 2: Invert
$\lambda=h/p$ is divided by $\sqrt3$, option (A).

Final Answer:
New to old wavelength is $1/\sqrt3$, option (A). \[ \boxed{\dfrac1{\sqrt3}} \]