Question

A proton, an electron, and an alpha particle have the same energies. Their de-Broglie wavelengths will be compared as:

• A. \( \lambda_e>\lambda_\alpha>\lambda_p \)
• B. \( \lambda_\alpha<\lambda_p<\lambda_e \)
• C. \( \lambda_p<\lambda_e<\lambda_\alpha \)
• D. \( \lambda_p>\lambda_e>\lambda_\alpha \)

Hint:

For particles with the same energy, the de-Broglie wavelength is inversely proportional to the square root of their mass.

Answer 1

The Correct Option is B

The de-Broglie wavelength is given by: \[ \lambda = \frac{h}{\sqrt{2mE}}, \] where \( h \) is Planck's constant, \( m \) is the mass of the particle, and \( E \) is its energy. For particles with the same energy, \( \lambda \propto \frac{1}{\sqrt{m}} \). The masses of the particles are: Electron (\( m_e \)): least mass, Proton (\( m_p \)): greater mass, Alpha particle (\( m_\alpha = 4m_p \)): greatest mass. Since \( \lambda \propto \frac{1}{\sqrt{m}} \), the order of the wavelengths is: \[ \lambda_e>\lambda_p>\lambda_\alpha. \] \[ \boxed{\lambda_\alpha<\lambda_p<\lambda_e} \]