de-Broglie wavelength of electrons of kinetic energy E is `lambda.` What will be its value if kinetic energy of electrons is made 4E ?

Find the de-Broglie wavelength of an electron with kinetic energy of 120 eV.

Find the de-Broglie wavelength of an electron with kinetic energy of 120 eV.

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.

The de broglie wavelength of electron moving with kinetic energy of 144 eV is nearly

The electron is moving with a velocity c/3 . The de-Broglie wavelength of electron is observed to be equal to a moving photon. The ratio of kinetic energy of electron to that energy of photon is found to be a:6. Find the value of a.

A photon and an electron moving with a velocity v have the same de broglie wavelength . Then the ratio of the kinetic energy of the electron to the kinetic energy of the photon is [C is the speed of light ]

The de-Broglie wavelength associated with an electron having a kinetic energy of 10 eV is

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 ?

Find the de Broglie wavelength of an electron if its kinetic energy is is orbiting in first excited state of hydrogen.