Dual Nature of ...

A proton and an electron are accelerated by the same potential difference. Let $$\lambda_{e}$$ and $$\lambda_{p}$$ denote the de Broglie wavelength of the electron and the proton respectively, then

An electron of mass $$9.1\times 10^{-31}$$kg and charge $$1.6\times 10^{-19}$$C is accelerated through a potential difference of 'V' volt. The de Broglie wavelength $$(\lambda )$$ associated with the electron is

The energy that should be added to an electron to reduce its de-Broglie wavelength from 1nm to 0.5nm is

If an electron and a proton have the same kinetic energy, the ratio of the de Broglie wavelengths of proton and electron would approximately be :

A particle having a de Broglie wavelength of 1.0 $$A^{0}$$ is associated with a momentum of (given $$h = 6.6 \times 10^{-34}$$ Js)

The de Broglie wavelength of a molecule of thermal energy KT (K is Boltzmann constant and T is absolute temperature) is given by :

The magnitude of the De-Broglie wavelength ($$\lambda$$) of an electron (e),proton(p),neutron (n) and $$\alpha $$ - particle ($$\alpha $$ ) all having the same energy of MeV, in the increasing order will follow the sequence:

The de Broglie wavelength of an electron having 80 eV of energy is nearly 

A particle of mass $$10^{-31}$$ kg is moving with a velocity equal to $$10^{5} ms^{-1}$$. The wavelength of the particle is equal to:

The de-Broglie wavelength of a particle moving with a velocity $$2.25 \times 10^{8}\ m/s$$ is equal to the wavelength of photon. The ratio of kinetic energy of the particle to the energy of the photon is :