An electron, a doubly ionized helium ion (\(\mathrm{He^{++}}\)) 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 :
Show Hint
- \(\lambda_e \gt \lambda_{He^{++}} \gt \lambda_p\)
- \(\lambda_e \lt \lambda_p \lt \lambda_{He^{++}}\)
- \(\lambda_e \gt \lambda_p \gt \lambda_{He^{++}}\)
- \(\lambda_e \lt \lambda_{He^{++}} = \lambda_p\)
The Correct Option is C
Solution and Explanation
With the same K, \(\lambda\propto\frac{1}{\sqrt m}\).
Masses: \(m_e \lt m_p \lt m_{He^{++}}\) (\(m_{He}\approx 4m_p\)).
Therefore \(\lambda_e \gt \lambda_p \gt \lambda_{He^{++}}\).
Answer: (3) \(\lambda_e \gt \lambda_p \gt \lambda_{He^{++}}\)
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