Question

An electron, a proton, an \(\alpha\)-particle and a hydrogen atom are moving with the same kinetic energy. The associated de-Broglie wavelength will be longest for --

• A. proton
• B. electron
• C. hydrogen atom
• D. \(\alpha\)-particle

Hint:

De-Broglie Wavelength: Lighter particles have longer wavelengths for the same energy. Think of it as smaller things having "wavier" properties.

Answer 1

The Correct Option is B

The de-Broglie wavelength (\(\lambda\)) is related to kinetic energy (K) and mass (m) by the equation: \[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} \] where h is Planck's constant. Since the kinetic energy K is the same for all particles, the wavelength \(\lambda\) is inversely proportional to the square root of the mass (\( \lambda \propto \frac{1}{\sqrt{m}} \)).
To have the longest wavelength, the particle must have the smallest mass. Comparing the masses:
Mass of electron (\(m_e\))<Mass of proton (\(m_p\))<Mass of hydrogen atom (\(m_H \approx m_p\))<Mass of \(\alpha\)-particle (\(m_\alpha \approx 4m_p\)). The electron has the smallest mass, and therefore it will have the longest de-Broglie wavelength.