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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.
Answer & Solution
Correct Answer
(D) \(\lambda_{e}>\lambda_{P}>\lambda_{H e^{++}}\)
Step-by-step Solution
Detailed explanation
\(\lambda=\frac{ h }{ P }=\frac{ h }{\sqrt{2 m ( KE )}}\)
\(\lambda \propto \frac{1}{\sqrt{ m }} \Rightarrow \lambda=\frac{ C }{\sqrt{ m }}\)
\(m _{ He ^{++}}> m _{ P }> m _{ e }\)
\(\therefore \lambda_{ He ^{++}}<\lambda_{ P }<\lambda_{ e }\)
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