A particle moving with kinetic energy E has de Broglie wavelength lambda .If energy Delta E is added to its energy the wavelength become (lambda)/(2) value of Delta E is

de-Broglie wavelength lambda is

The de Broglie wavelength lambda of a particle

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

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

An electron is having a kinetic energy of 50 eV. Its de broglie wavelength is

An electron is having a kinetic energy of 50 eV. Its de-Broglie wavelength is

Kinetic energy of the particle is E and it's De-Broglie wavelength is lambda . On increasing it's KE by delta E , it's new De-Broglie wavelength becomes lambda/2 . Then delta E is

If the kinetic energy of a particle is doubled, de Broglie wavelength becomes: