Two particles move at right angle to each other. Their de Broglie wavelenghts are `lambda_(1)` and `lambda_(2)` respectively. The particles suffer perfectly inelastic collision. The de Broglie wavelenght `lambda`, of the final particle, is given by:
Two particles move at right angle to each other.Their de Broglie wavelengths are lambda_(1) and lambda_(2) respectively.The particles suffer perfectly inelastic collision.The de Broglie wavelength lambda of the final particle is given by :
The de Broglie wavelength lambda of a particle
Two particles having de-Broglie wavelengths lambda_(1) and lambda_(2) , while moving along mutually perpendicular directions, undergo perfectly inelastic collision. The de-Broglie wavelength lambda , of the final particle is
The de-Broglie wavelength associated with a particle of momentum p is given as :
Two identical non-re,aticivistic particles move at right angle to each other. Possesing de Broglie wavelength lambda_(1) and lambda_(2) Find the Broglie wavelength of each particle in the frame of their centre of inertia.
The de Broglie wavelength (lambda) of a particle is related to its kinetic energy E as
Two identical non-relativitic partcles A and B move at right angles to each othre, processing de Broglie wavelengths lamda_1 and lamda_2 , respectively. The de Broglie wavelength of each particle in their centre of mass frame of reference is
Two particles A and B of de-broglie wavelength lambda_(1) and lambda_(2) combine to from a particle C. The process conserves momentum. Find the de-Broglie wavelength of the particle C. (The motion is one dimensional).
An electron, a neutron and an alpha particle have same kinetic energy and their de-Broglie wavelength are lambda_e, lambda_n and lambda_(alpha) respectively. Which statement is correct about their de-Broglie wavelengths?
