An electron, an `alpha`-particle, and a photon have the same kinetic energy. Which of these particles has the shortest, de -broglie wavelenght?

Options

A. Deuteron
B. Cl-particle
C. Proton
D. Electron

Answer

B

Explanation

The de Broglie wavelength of a charged particle is ` lamda = h/(sqrt(2mK))`
where m is the mass and K is the kinetic energy of the charged particle.
As all thte four given charged particles have same kinetic energy .
`:. lamda prop 1/sqrtm`
As `m_("electron") lt m_("proton") lt m_("deuteron") lt m_(alpha- " particle") `
`m_("electron") gt m_("proton") gt m_("deuteron") gt m_(alpha- " particle") `
Hence, `alpha` - particle has the shortest de Broglie wavelength.