The de Broglie wavelength of an electron, a proton, and a alpha particle are represented by \( \lambda_e, \lambda_p, \lambda_\alpha \) respectively. If kinetic energy of electron, proton and alpha particle is same, then which of the following is correct.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \lambda_\alpha \lt \lambda_p \lt \lambda_e \)
The de Broglie wavelength is \( \lambda = h/p \), and a particle of mass \( m \) and kinetic energy \( K \) has momentum \( p = \sqrt{2mK} \).
\( \displaystyle \lambda = \frac{h}{\sqrt{2mK}} \)
For the same kinetic energy, \( \lambda \propto 1/\sqrt{m} \): the heavier the particle, the shorter its wavelength.
\( m_e \lt m_p \lt m_\alpha \;(m_\alpha \approx 4m_p) \)
\( \lambda_\alpha \lt \lambda_p \lt \lambda_e \)
Note: This question is from the NCERT Exemplar. NTA sets numerical questions from the Exemplar too, so do not ignore it while practising numericals.
NCERT Exemplar:
Dual Nature of Radiation and Matter – …
