The Kinetic Energy of an electron having de Broglie wavelength (\( \lambda \)) triple, then the de Broglie wavelength associated with it becomes:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{\lambda}{\sqrt{3}} \)
The de Broglie wavelength of a particle of mass \( m \) and kinetic energy \( K \) is
\( \displaystyle \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} \;\Rightarrow\; \lambda \propto \frac{1}{\sqrt{K}} \)
So the wavelength is inversely proportional to the square root of the kinetic energy. When the kinetic energy triples, \( K' = 3K \):
\( \displaystyle \frac{\lambda'}{\lambda} = \sqrt{\frac{K}{K'}} = \frac{1}{\sqrt{3}} \;\Rightarrow\; \lambda' = \frac{\lambda}{\sqrt{3}} \)