Question

Which of these particles (having the same kinetic energy) has the largest de-Broglie wavelength?

• A. Electron
• B. Alpha particle
• C. Proton
• D. Neutron

Answer 1

Answer: (A)

de-Broglie wavelength $=\lambda =\frac{h}{mv}=\frac{h}{\sqrt{2mE}}$
$E$ is same for all, so $\lambda \propto \frac{1}{\sqrt{m}}$
Hence, de-Broglie waveiength will be maximum for particle with lesser mass. Mass of the given particles in increasing order are given as
$m_e<m_p<m_n<m_{\alpha }$
Thus, de-Broglie wavelength will be maximum for electron.