AP EAMCET · Chemistry · Structure of Atom
If the kinetic energy of a particle is reduced to half, de-Broglie wavelength becomes
- A \(2\) times
- B \(\frac{1}{\sqrt{2}}\) times
- C \(4\) times
- D \(\sqrt 2\) times
Answer & Solution
Correct Answer
(D) \(\sqrt 2\) times
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { } \lambda \propto \frac{1}{\sqrt{\mathrm{K} . E}} \Rightarrow \frac{\lambda_1}{\lambda_2}=\sqrt{\frac{(\mathrm{KE})_2}{(\mathrm{KE})_1}} \\ & \mathrm{KE}_2=\frac{(\mathrm{KE})_1}{2} \\ & \text { Thus, } \frac{\lambda_1}{\lambda_2}=\sqrt{\frac{1}{2}}…
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