The de Broglie wavelength `(lambda)` of a particle is related to its kinetic energy E as

The de-Broglie wavelength lambda . of a particle isrelated to its kinetic energy E_K as

In what way de - Broglie wavelength lambda of a particle is related to K.E?

Identify the graph depicting the variation of the de Broglie wavelength lambda of an electron with its kinetic energy K.

The de - Broglie wavelength lambda associated with an electron having kinetic energy E is given by the expression

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

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

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

Assertiom: The de-broglie wavelength of a neutrons when its kinetic energy is k is lambda . Its wavelength is 2lambda when its kinetic energy is 4k . Reason : The de-broglie wavelength lambda is proportional to square root of the kinetic energy.