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JEE Mains · Physics · STD 12 - 11. Dual nature of radiation and matter
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
- A \(\lambda_{ e }<\lambda_{ P }<\lambda_{ He ^{++}}\)
- B \(\lambda_{e}<\lambda_{H e^{++}}=\lambda_{P}\)
- C \(\lambda_{e}>\lambda_{H e^{++}}>\lambda_{P}\)
- D \(\lambda_{e}>\lambda_{P}>\lambda_{H e^{++}}\)
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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