Advertisements
Advertisements
प्रश्न
An electron is accelerated from rest through a potential V. Obtain the expression for the de-Broglie wavelength associated with it ?
Advertisements
उत्तर
Energy of electron, E = eV
Momentum,
\[p = \sqrt{2mE} = \sqrt{2meV}\]
de-Broglie wavelength,
\[\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}}\]
∴\[\lambda = \frac{h}{\sqrt{2meV}}\]
संबंधित प्रश्न
A proton and an α-particle have the same de-Broglie wavelength. Determine the ratio of their accelerating potentials
State how de-Broglie wavelength (`lambda`) of moving particles varies with their linear momentum (p).
A proton and an electron are accelerated by the same potential difference. Let λe and λpdenote the de Broglie wavelengths of the electron and the proton, respectively.
What is meant by diffraction of light?
If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is:
When the displacement of a particle executing simple harmonic motion is half of its amplitude, the ratio of its kinetic energy to potential energy is:
Find the de-Broglie Wavelength for an electron moving at the speed of 5.0 × 106 m/s. (mass of electron is 9.1 × 10-31)
For what k.e of neutron will the associated de-Broglie wavelength be 1.40 × 10-10 m? mass of neutron= 1.675 × 10-27 kg, h = 6.63 × 10-34 J
The work function of a metal is 4.50 eV. Find the frequency of light to be used to eject electrons from the metal surface with a maximum kinetic energy of 6.06 × 10−19 J.
What is meant by “Dual nature of matter”?
