ExamBro
ExamBro
JEE Mains · Physics · STD 12 - Dual nature of radiation and matter
  • JEE Mains 2020
  • Medium
  • MCQ

An electron, a doubly ionized helium ion \(\left( He ^{++}\right)\) and a proton are having the same kinetic energy. The relation between their respective de\(-\)Broglie wavelengths \(\lambda_{ e }, \lambda_{ He ^{++}}\) and \(\lambda_{ P }\) is

Pick an option to check your answer.
Verified Solution

Answer & Solution

Correct Answer

(D) \(\lambda_{e}>\lambda_{P}>\lambda_{H e^{++}}\)

Step-by-step Solution

Detailed explanation

\(\lambda=\frac{ h }{ P }=\frac{ h }{\sqrt{2 m ( KE )}}\)

\(\lambda \propto \frac{1}{\sqrt{ m }} \Rightarrow \lambda=\frac{ C }{\sqrt{ m }}\)

\(m _{ He ^{++}}> m _{ P }> m _{ e }\)

\(\therefore \lambda_{ He ^{++}}<\lambda_{ P }<\lambda_{ e }\)

Same subject
Explore more questions on app