Question

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

पर्याय

• λ_e > λ_α > λ_p

• λ_α < λ_p < λ_e

• λ_p < λ_e < λ_α

• λ_p > λ_e > λ_α

Answer 1

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

Explanation:

Mass of electron (m_e) is the lightest (approx. 1/1836 amu). Mass of proton (m_p) is middle (approx. 1 amu). The mass of an alpha particle is the heaviest (approx. 4 amu).

Hence,

m_α > m_p > m_e

For particles with the same kinetic energy, the de-Broglie wavelength is:

λ = `h/(sqrt (2 m K))`

⇒ λ ∝ `1/sqrt m`

Since λ is inversely proportional to the square root of mass, the order reverses:

λ_α < λ_p < λ_e