A free particle with initial kinetic energy E, de-Broglie wavelength `lambda`, enters a region where in it has a potential V, what is the new de-Broglie wavelength?

The de-Broglie wavelength lambda

de-Broglie wavelength lambda is

A particle is moving with kinetic energy E. It's de Broglie wavelength is—

The de Broglie wavelength lambda of a particle

De Broglie wavelength lambda is proportional to

De Broglie wavelength lambda is proportional to

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