De broglie wavelength `(lambda)` k of a particle is related to its momentum (p) as `lambda=(h)/(p )` where h is planck s constant or ` lambda=(h) /(mv)` `rarr lambda` is inversely proportional to impluse (F)
Topper's Solved these Questions
MHTCET 2015
MHTCET PREVIOUS YEAR PAPERS AND PRACTICE PAPERS|Exercise Chemistry|1 Videos
MHTCET 2014
MHTCET PREVIOUS YEAR PAPERS AND PRACTICE PAPERS|Exercise Physics|45 Videos
MHTCET 2016
MHTCET PREVIOUS YEAR PAPERS AND PRACTICE PAPERS|Exercise Physics|50 Videos
Similar Questions
Explore conceptually related problems
The de-Broglie wavelength lamda
de-Broglie wavelength lambda is
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 :
A particle of charge q mass m and energy E has de-Broglie wave length lambda .For a particle of charge 2q mass 2m and energy 2E the de-Broglie wavelength is
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
De Broglie wavelength lambda is proportional to
The de-Broglie wavelength of a particle having kinetic energy E is lamda . How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to 75% of the initial value ?
The de Broglie wavelength is given by
A particle moving with kinetic energy E_(1) has de Broglie wavelength lambda_(1) . Another particle of same mass having kinetic energy E_(2) has de Broglie wavelength lambda_(2). lambda_(2)= 3 lambda_(2) . Then E_(2) - E_(1) is equal to
The de Broglie wavelength associated with particle is
MHTCET PREVIOUS YEAR PAPERS AND PRACTICE PAPERS-MHTCET 2015-Chemistry
