A particle of mass 2.21 10⁻²⁸ ~kg has a de Broglie wavelength exactly equal to the wavelength of an electromagnetic wave propagating in a vacuum with a frequency of 10¹⁴ ~Hz . The velocity of the particle is (Take c = 3 10^8 ~m/s , h = 6.63 10⁻³⁴ ~J.s )

Step-by-step solution

The wavelength of the electromagnetic wave is given by: = c = 3 10^8 10¹⁴ = 3 10⁻⁶ ~m The de Broglie wavelength of the particle is equal to this wavelength: _ dB = = 3 10⁻⁶ ~m Using the de Broglie relation _ dB = h mv , the velocity of the particle is: v = h m _ dB Substituting the given values: v = 6.63 10⁻³⁴ 2.21 10⁻²⁸ 3 10⁻⁶ v = 6.63 10⁻³⁴ 6.63 10⁻³⁴ = 1 ~m/s Answer: 1 ~m/s