A sub-atomic particle of mass 3.315 10⁻²⁸ ~kg is moving such that its kinetic energy is exactly equal to the energy of a photon of frequency 10¹⁴ ~Hz . Under the matter wave consideration, the particle will behave closely like ( h =6.63 10⁻³⁴ ~J . s )

Step-by-step solution

The energy of the photon is given by Planck's equation: E = h = 6.63 10⁻³⁴ 10¹⁴ = 6.63 10⁻²⁰ ~J The kinetic energy K of the particle is equal to the photon's energy: K = 6.63 10⁻²⁰ ~J The momentum of the particle is: p = 2mK Substituting the values of mass and kinetic energy: p = 2 3.315 10⁻²⁸ 6.63 10⁻²⁰ p = 6.63 10⁻²⁸ 6.63 10⁻²⁰ = 6.63 10⁻²⁴ ~kg m/s The de Broglie wavelength of the particle is: = h p = 6.63 10⁻³⁴ 6.63 10⁻²⁴ = 10⁻¹⁰ ~m A wavelength of 10⁻¹⁰ ~m ( 1 ~ A ) falls in the X-ray region of the electromagne