A particle of mass 1 10⁻³⁰ ~kg and electric change 1.6 10⁻¹⁹ C has de-Broglie wavelength 660 ~nm . Then kinetic energy of this particle is (Planck's constant, h=6.6 10⁻³⁴ ~J - s )

Step-by-step solution

de-Broglie wavelength is given by =h / p Momentum of particle, aligned & p= h = 6.6 10⁻³⁴ 6.60 10⁻⁹ =1 10⁻²⁷ ~kg ~m / s & m v=1 10⁻²⁷ & or v= 1 10⁻²⁷ 1 10⁻³⁰ v=10^3 ~m / s & aligned Kinetic energy of particle is aligned K & = 1 2 m v^2 & = 1 2 1 10⁻³⁰ (10^3 )^2 & =0.5 10⁻²⁴ ~J aligned As, 1 eV =1.6 10⁻¹⁹ ~J Kinetic energy of particle, aligned K & = 0.5 10⁻²⁴ 1.6 10⁻¹⁹ eV =0.3125 10⁻⁵ & =3125 10⁻⁶ eV 3.1 10⁻⁶ eV aligned