JEE Mains · Physics · STD 12 - 11. Dual nature of radiation and matter
The kinetic energy of an electron, \(\alpha\)-particle and a proton are given as \(4K,2K\) and \(K\) respectively. The de-Broglie wavelength associated with electron ( \(\lambda e), \alpha\)-particle \((\lambda \alpha)\) and the proton \((\lambda p)\) are as follows:
- A \(\lambda \alpha=\lambda p < \lambda e\)
- B \(\lambda \alpha > \lambda p > \lambda e\)
- C \(\lambda \alpha < \lambda p < \lambda e\)
- D \(\lambda \alpha=\lambda p>\lambda e\)
Answer & Solution
Correct Answer
(C) \(\lambda \alpha < \lambda p < \lambda e\)
Step-by-step Solution
Detailed explanation
\(Electron\) \(Alpha\) \(Proton\) \(Mass:\) \(\frac{m}{1840}\) \(4m\) \(m\) \(Charge:\) \(e\) \(2e\) \(e\) \(Kinetic energy:\) \(4K\) \(2K\) \(K\) \(\lambda=\frac{h}{\sqrt{2 mK }}\) \(\frac{ h }{\sqrt{2 \cdot \frac{ m }{1840} \cdot 4 K }}\) \(\frac{h}{\sqrt{2.4 m.2K }}\)…
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