Question

A proton, a deuteron, an electron, and an \(\alpha\)-particle have the same energy. Their deBroglie wavelengths are \( \lambda_p \), \( \lambda_d \), \( \lambda_e \), and \( \lambda_{\alpha} \), respectively. Which of the following is correct?

• A. \( \lambda_{\infty}<\lambda_d<\lambda_p = \lambda_e \)
• B. \( \lambda_{\infty}<\lambda_d<\lambda_p<\lambda_e \)
• C. \( \lambda_e<\lambda_p = \lambda_d<\lambda_{\infty} \)
• D. \( \lambda_e = \lambda_p = \lambda_d<\lambda_{\infty} \)

Hint:

For particles with the same energy, the deBroglie wavelength decreases as the particle's mass increases.

Answer 1

The Correct Option is A

The deBroglie wavelength is given by the equation: \[ \lambda = \frac{h}{\sqrt{2mE}} \] Since the proton, deuteron, and electron have the same energy, their deBroglie wavelengths depend on their masses. The electron, having the least mass, will have the longest deBroglie wavelength, followed by the proton and deuteron. The \(\alpha\)-particle, having the highest mass, will have the shortest wavelength. Thus, the correct order of wavelengths is \( \lambda_{\infty}<\lambda_d<\lambda_p = \lambda_e \).