JEE Mains · Physics · STD 12 - Dual nature of radiation and matter
- JEE Mains 2023
- Medium
- MCQ
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:
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(\lambda \alpha < \lambda p < \lambda e\)
Step-by-step Solution
Detailed explanation
| \(Electron\) | \(Alpha\) | \(Proton\) | |
| \(Mass:\) | \(\frac{m}{1840}\) | \(4m\) | \(m\) |
| \(Charge:\) | \(e\) | \(2e\) | \(e\) |
| \(Kinetic energy:\) | \(4K\) | \(2K\) | \(K\) |
| \(\lambda=\frac{h}{\sqrt{2 mK }}\) | \(\frac{ h }{\sqrt{2 \cdot \frac{ m }{1840} \cdot 4 K }}\) | \(\frac{h}{\sqrt{2.4 m.2K }}\) | \(\frac{h}{\sqrt{2 mK }}\) |
\(\lambda_\alpha < \lambda_p < \lambda_e\)
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