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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:

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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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