WBJEE · Chemistry · Structure of Atom
The de-Broglie wavelength \((\lambda)\) for electron (e), proton \((p)\) and \(\mathrm{He}^{2+}\) ion \((\alpha)\) are in the following order. (Speed of e, p and \(\alpha\) are the same)
- A \(\alpha>p>e\)
- B \(e>p>\alpha\)
- C \(e>\alpha>p\)
- D \(\alpha < p>e\)
Answer & Solution
Correct Answer
(B) \(e>p>\alpha\)
Step-by-step Solution
Detailed explanation
\(\lambda=\frac{\mathrm{h}}{\mathrm{mv}}\) \(\because\) speed are same \(\begin{aligned} &\therefore \lambda \propto \frac{1}{\mathrm{~m}} \\ &\therefore \lambda_e>\lambda_p>\lambda_\alpha \text { as } \quad \mathrm{m}_\alpha>\mathrm{m}_{\mathrm{p}}>\mathrm{m}_e \end{aligned}\)
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