Two particles of de-broglie wavelength `lamda_(x)` and `lamda_(y)` are moving in opposite direction. Find debroglie wavelength after perfectly inelastic collision:
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
A particle ‘P’ is formed due to a completely inelastic collision of particles ‘x’ and ‘y’ having de-Broglie wavelength lambda_(x) and lambda_(y) respectively. If x and y were moving in opposite directions, then the de-Broglie wavelength of ‘P’ is:
The de Broglie wavelength lambda of a particle
The de Broglie wavelength associated with particle 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).
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 :
Assertion : Two particles of de Broglie wavelength lamda_1 and lamda_2 combine to form a third particle of wavelength lamda_3 . Then , lamda_1 +lamda_2 =lamda_3 . Reason : If p_1,p_2 and p_3 are momenta of the three particles , respectively, then p_1+p_2=p_3 .
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
